IB Mathematics: Analysis and Approaches Quiz: Normal Distribution Extensions
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Normal Distribution ExtensionsQuestion 1 of 20

A manufacturer claims that their light bulbs have a mean lifetime of 1000 hours. The lifetimes are not normally distributed, but have a known standard deviation of 120 hours. A consumer group tests a sample of 36 bulbs. Using the Central Limit Theorem, what is the probability that the sample mean lifetime is less than 950 hours?

0.0062
0.0228
0.3385
It cannot be determined without knowing the population distribution.
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Normal Distribution Extensions

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Question 1

A manufacturer claims that their light bulbs have a mean lifetime of 1000 hours. The lifetimes are not normally distributed, but have a known standard deviation of 120 hours. A consumer group tests a sample of 36 bulbs. Using the Central Limit Theorem, what is the probability that the sample mean lifetime is less than 950 hours?

  1. 0.0062 (correct answer)
  2. 0.0228
  3. 0.3385
  4. It cannot be determined without knowing the population distribution.
Explanation: We are given μ=1000\mu = 1000, σ=120\sigma = 120, and n=36n=36. The population is not normal.
However, since the sample size n=36n=36 is large enough (n≥30n \ge 30), we can apply the Central Limit Theorem. The CLT states that the sampling distribution of the sample mean Xˉ\bar{X} is approximately normal.
The mean of this distribution is μXˉ=μ=1000\mu_{\bar{X}} = \mu = 1000.
The standard deviation of this distribution (standard error) is σXˉ=σn=12036=1206=20\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{120}{\sqrt{36}} = \frac{120}{6} = 20.
So, Xˉ≈N(1000,202)\bar{X} \approx N(1000, 20^2).
We want to find P(Xˉ<950)P(\bar{X} < 950).
Standardize the value: Z=950−100020=−5020=−2.5Z = \frac{950 - 1000}{20} = \frac{-50}{20} = -2.5.
We need P(Z<−2.5)P(Z < -2.5). Using a calculator or standard normal table, this probability is approximately 0.0062.\
  • B: 0.0228 corresponds to a Z-score of -2. This would occur if the standard error was miscalculated as 25 (e.g. 1000−950=50,50/2=251000-950=50, 50/2 = 25).
  • C: 0.3385 is the result of forgetting to use the standard error and using the population standard deviation instead: Z=950−1000120≈−0.4167Z = \frac{950 - 1000}{120} \approx -0.4167, for which P(Z<−0.4167)≈0.3385P(Z < -0.4167) \approx 0.3385.
  • D: It cannot be determined... This is incorrect because the Central Limit Theorem allows us to approximate the probability due to the large sample size.

Question 2

The lifetime of a particular type of battery is a random variable with a mean of 400 hours and a standard deviation of 50 hours. The distribution of the lifetime is unknown. A random sample of 100 batteries is selected. Which of the following allows for the calculation of the approximate probability that the sample mean lifetime is greater than 410 hours?

  1. The assumption that the population distribution is normal.
  2. The Central Limit Theorem. (correct answer)
  3. The Law of Large Numbers.
  4. The use of a continuity correction.
Explanation: The problem asks for the probability related to the sample mean Xˉ\bar{X}. The distribution of the original population is explicitly stated as unknown. Therefore, we cannot assume it is normal.
However, the sample size n=100n=100 is large (typically n≥30n \ge 30 is considered sufficient).
The Central Limit Theorem (CLT) states that for a sufficiently large sample size, the sampling distribution of the sample mean Xˉ\bar{X} will be approximately normal, regardless of the shape of the population distribution. This is precisely the theorem that justifies approximating the distribution of Xˉ\bar{X} with a normal distribution to calculate the required probability.\
  • A: The assumption that the population distribution is normal. If this were true, the distribution of Xˉ\bar{X} would be exactly normal, not approximately. The CLT allows us to proceed without this assumption.
  • C: The Law of Large Numbers. This law states that as the sample size grows, the sample mean Xˉ\bar{X} converges to the population mean μ\mu. It describes the long-term behavior of the mean, but not its sampling distribution for a finite sample.
  • D: The use of a continuity correction. This is used when approximating a discrete probability distribution (like Binomial or Poisson) with a continuous one (like Normal). Battery lifetime is a continuous variable, so this correction is not applicable.

Question 3

A machine produces bolts with lengths that are normally distributed with a standard deviation of 0.1 mm. The mean length, μ\mu, is adjustable. For a batch to be accepted, the probability that the mean length of a sample of 25 bolts is within 0.05 mm of μ\mu must be at least 0.99. What is the maximum possible value of the standard deviation of the population for this condition to be met?

  1. 0.0194 mm
  2. 0.0970 mm (correct answer)
  3. 0.4852 mm
  4. 2.5758 mm
Explanation: Let XX be the length of a bolt, X∼N(μ,σ2)X \sim N(\mu, \sigma^2). We are given n=25n=25. The sample mean Xˉ\bar{X} is distributed as Xˉ∼N(μ,σ2/25)\bar{X} \sim N(\mu, \sigma^2/25). The standard error is σXˉ=σ/25=σ/5\sigma_{\bar{X}} = \sigma/\sqrt{25} = \sigma/5.
We require the probability that the sample mean is 'within 0.05 mm of μ\mu' to be at least 0.99. This is written as P(∣Xˉ−μ∣≤0.05)≥0.99P(|\bar{X} - \mu| \le 0.05) \ge 0.99.
This is equivalent to P(−0.05≤Xˉ−μ≤0.05)≥0.99P(-0.05 \le \bar{X} - \mu \le 0.05) \ge 0.99.
To standardize, we divide by the standard error σ/5\sigma/5:
P(−0.05σ/5≤Z≤0.05σ/5)≥0.99P(\frac{-0.05}{\sigma/5} \le Z \le \frac{0.05}{\sigma/5}) \ge 0.99.
Let zc=0.05σ/5=0.25σz_c = \frac{0.05}{\sigma/5} = \frac{0.25}{\sigma}. We need P(−zc≤Z≤zc)≥0.99P(-z_c \le Z \le z_c) \ge 0.99.
This means the area in the two tails is at most 1−0.99=0.011 - 0.99 = 0.01. The area in the upper tail is at most 0.005. So P(Z>zc)≤0.005P(Z > z_c) \le 0.005.
This implies P(Z<zc)≥0.995P(Z < z_c) \ge 0.995.
Using the inverse normal function, the critical z-score for a cumulative probability of 0.995 is zcrit≈2.5758z_{crit} \approx 2.5758.
So we must have zc≥2.5758z_c \ge 2.5758.
0.25σ≥2.5758\frac{0.25}{\sigma} \ge 2.5758.
σ≤0.252.5758≈0.09705\sigma \le \frac{0.25}{2.5758} \approx 0.09705.
The maximum possible value for the standard deviation σ\sigma is approximately 0.0970 mm.\
  • A: 0.0194 is 0.05/2.57580.05 / 2.5758, from forgetting to use the standard error σ/5\sigma/5.
  • C: 0.4852 might come from an algebraic error, e.g., σ≤0.25×1.96\sigma \le 0.25 \times 1.96 using the wrong Z-score for 95% confidence.
  • D: 2.5758 is the Z-score itself, not the standard deviation.

Question 4

The weights of apples from a certain orchard are normally distributed with a mean of 150g and a standard deviation of 20g. A random sample of 16 apples is taken. What is the probability that the mean weight of the apples in the sample is less than 145g?

  1. 0.0228
  2. 0.1587 (correct answer)
  3. 0.4013
  4. 0.8413
Explanation: Let XX be the weight of a single apple. We have X∼N(150,202)X \sim N(150, 20^2).
For a sample of size n=16n=16, the distribution of the sample mean Xˉ\bar{X} is given by Xˉ∼N(μ,σ2n)\bar{X} \sim N(\mu, \frac{\sigma^2}{n}).
So, Xˉ∼N(150,20216)\bar{X} \sim N(150, \frac{20^2}{16}).
The standard deviation of the sample mean (the standard error) is σXˉ=σn=2016=204=5\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{16}} = \frac{20}{4} = 5.
Thus, Xˉ∼N(150,52)\bar{X} \sim N(150, 5^2).
We need to find P(Xˉ<145)P(\bar{X} < 145).
Standardizing the value: Z=145−1505=−55=−1Z = \frac{145 - 150}{5} = \frac{-5}{5} = -1.
We need P(Z<−1)P(Z < -1). From standard normal tables or a calculator, this probability is approximately 0.1587.\
  • A: 0.0228 corresponds to a Z-score of -2, which might result from a calculation error for the standard error, such as σXˉ=2.5\sigma_{\bar{X}} = 2.5.
  • C: 0.4013 is the result of forgetting to divide the standard deviation by n\sqrt{n} and using σ=20\sigma = 20 instead of σXˉ=5\sigma_{\bar{X}}=5. This gives Z=145−15020=−0.25Z = \frac{145-150}{20} = -0.25, and P(Z<−0.25)≈0.4013P(Z < -0.25) \approx 0.4013.
  • D: 0.8413 is P(Z>−1)P(Z > -1), which would answer the question 'what is the probability the mean weight is more than 145g?'.

Question 5

The weights of two breeds of dog, A and B, are independent and normally distributed. For breed A, the mean weight is 30 kg and 10% of dogs weigh more than 35 kg. For breed B, the mean weight is 25 kg and the standard deviation is 3 kg. What is the probability that the total weight of two randomly selected dogs, one from each breed, exceeds 60 kg?

  1. 0.121
  2. 0.155 (correct answer)
  3. 0.196
  4. 0.234
Explanation: Let XX be the weight of a dog of breed A and YY be the weight of a dog of breed B.
Step 1: Find the standard deviation of breed A, σA\sigma_A. We are given X∼N(30,σA2)X \sim N(30, \sigma_A^2) and P(X>35)=0.10P(X > 35) = 0.10. Standardizing, P(Z>35−30σA)=0.10P(Z > \frac{35-30}{\sigma_A}) = 0.10. This means P(Z<5σA)=0.90P(Z < \frac{5}{\sigma_A}) = 0.90. Using the inverse normal function, the z-score corresponding to a cumulative probability of 0.90 is approximately 1.2816. So, 5σA≈1.2816  ⟹  σA≈51.2816≈3.901\frac{5}{\sigma_A} \approx 1.2816 \implies \sigma_A \approx \frac{5}{1.2816} \approx 3.901 kg.
Step 2: Define the total weight T=X+YT = X + Y. We have Y∼N(25,32)Y \sim N(25, 3^2). The distribution of TT is also normal.
E(T)=E(X)+E(Y)=30+25=55E(T) = E(X) + E(Y) = 30 + 25 = 55 kg.
Var(T)=Var(X)+Var(Y)=(3.901)2+32≈15.218+9=24.218Var(T) = Var(X) + Var(Y) = (3.901)^2 + 3^2 \approx 15.218 + 9 = 24.218.
σT=24.218≈4.921\sigma_T = \sqrt{24.218} \approx 4.921 kg.
Step 3: Calculate P(T>60)P(T > 60). Standardizing, Z=60−554.921≈1.016Z = \frac{60 - 55}{4.921} \approx 1.016.
P(Z>1.016)=1−P(Z<1.016)≈1−0.8452=0.1548P(Z > 1.016) = 1 - P(Z < 1.016) \approx 1 - 0.8452 = 0.1548. This is approximately 0.155.\
  • A: 0.121 results from using an incorrect z-score for 10% tail, such as 1.645 (for a 5% tail).
  • C: 0.196 results from incorrectly assuming σA=35−30=5\sigma_A = 35 - 30 = 5.
  • D: 0.234 results from incorrectly adding standard deviations (σT=3.901+3\sigma_T = 3.901 + 3) instead of variances.

Question 6

In a large population, 40% of people have blood type A. A random sample of 200 people is selected. Using a normal approximation with continuity correction, which expression calculates the probability that between 75 and 90 people (inclusive) in the sample have blood type A?

  1. P(75<Y<90)P(75 < Y < 90), where Y∼N(80,48)Y \sim N(80, 48)
  2. P(74.5<Y<90.5)P(74.5 < Y < 90.5), where Y∼N(80,48)Y \sim N(80, 48) (correct answer)
  3. P(75.5<Y<89.5)P(75.5 < Y < 89.5), where Y∼N(80,48)Y \sim N(80, 48)
  4. P(74.5<Y<90.5)P(74.5 < Y < 90.5), where Y∼N(80,24)Y \sim N(80, 24)
Explanation: Let XX be the number of people with blood type A in the sample. This follows a binomial distribution X∼B(n,p)X \sim B(n, p) with n=200n=200 and p=0.4p=0.4.
To use a normal approximation, we find the mean and variance:
Mean: μ=np=200×0.4=80\mu = np = 200 \times 0.4 = 80.
Variance: σ2=np(1−p)=200×0.4×0.6=48\sigma^2 = np(1-p) = 200 \times 0.4 \times 0.6 = 48.
So we approximate XX with a normal distribution Y∼N(80,48)Y \sim N(80, 48).
The question asks for the probability that the number of people is between 75 and 90 inclusive, which is P(75≤X≤90)P(75 \le X \le 90).
When applying the continuity correction for an inclusive range [a,b][a, b], the interval becomes (a−0.5,b+0.5)(a-0.5, b+0.5).
Therefore, P(75≤X≤90)P(75 \le X \le 90) is approximated by P(74.5<Y<90.5)P(74.5 < Y < 90.5).
Combining these, the correct expression is P(74.5<Y<90.5)P(74.5 < Y < 90.5), where Y∼N(80,48)Y \sim N(80, 48).\
  • A: This expression does not use a continuity correction.
  • C: This expression uses an incorrect continuity correction, which would be appropriate for a strict inequality P(76<X<89)P(76 < X < 89).
  • D: This expression calculates the variance incorrectly, possibly by using np(p)np(p) instead of np(1−p)np(1-p).

Question 7

The number of calls received by a call centre in an hour follows a Poisson distribution with a mean of 40. Using a suitable approximation, find the probability that the call centre receives between 35 and 45 calls (inclusive) in a given hour.

  1. 0.109
  2. 0.523
  3. 0.571
  4. 0.615 (correct answer)
Explanation: Let XX be the number of calls. We are given X∼Po(40)X \sim Po(40). Since the mean λ=40\lambda = 40 is large (typically λ>10\lambda > 10), we can use a normal approximation Y∼N(λ,λ)Y \sim N(\lambda, \lambda), so Y∼N(40,40)Y \sim N(40, 40).
We want to find P(35≤X≤45)P(35 \le X \le 45). We must use a continuity correction.
P(35≤X≤45)P(35 \le X \le 45) is approximated by P(34.5<Y<45.5)P(34.5 < Y < 45.5).
The standard deviation is σ=40≈6.325\sigma = \sqrt{40} \approx 6.325.
We standardize the endpoints:
Z1=34.5−4040=−5.56.325≈−0.8696Z_1 = \frac{34.5 - 40}{\sqrt{40}} = \frac{-5.5}{6.325} \approx -0.8696
Z2=45.5−4040=5.56.325≈0.8696Z_2 = \frac{45.5 - 40}{\sqrt{40}} = \frac{5.5}{6.325} \approx 0.8696
We need P(−0.8696<Z<0.8696)=P(Z<0.8696)−P(Z<−0.8696)≈0.8077−0.1923=0.6154P(-0.8696 < Z < 0.8696) = P(Z < 0.8696) - P(Z < -0.8696) \approx 0.8077 - 0.1923 = 0.6154.\
  • A: 0.109 is the result of incorrectly using σ=40\sigma = 40 instead of σ=40\sigma = \sqrt{40}.
  • B: 0.523 is the result of using an incorrect continuity correction, such as P(35.5<Y<44.5)P(35.5 < Y < 44.5).
  • C: 0.571 is the result of not applying the continuity correction, calculating P(35<Y<45)P(35 < Y < 45).

Question 8

The weights of apples from a certain orchard are normally distributed with a mean of 150g and a standard deviation of 20g. A random sample of 16 apples is taken. What is the probability that the mean weight of the apples in the sample is less than 145g?

  1. 0.0228
  2. 0.1587 (correct answer)
  3. 0.4013
  4. 0.8413
Explanation: Let XX be the weight of a single apple. We have X∼N(150,202)X \sim N(150, 20^2).
For a sample of size n=16n=16, the distribution of the sample mean Xˉ\bar{X} is given by Xˉ∼N(μ,σ2n)\bar{X} \sim N(\mu, \frac{\sigma^2}{n}).
So, Xˉ∼N(150,20216)\bar{X} \sim N(150, \frac{20^2}{16}).
The standard deviation of the sample mean (the standard error) is σXˉ=σn=2016=204=5\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{16}} = \frac{20}{4} = 5.
Thus, Xˉ∼N(150,52)\bar{X} \sim N(150, 5^2).
We need to find P(Xˉ<145)P(\bar{X} < 145).
Standardizing the value: Z=145−1505=−55=−1Z = \frac{145 - 150}{5} = \frac{-5}{5} = -1.
We need P(Z<−1)P(Z < -1). From standard normal tables or a calculator, this probability is approximately 0.1587.\
  • A: 0.0228 corresponds to a Z-score of -2, which might result from a calculation error for the standard error, such as σXˉ=2.5\sigma_{\bar{X}} = 2.5.
  • C: 0.4013 is the result of forgetting to divide the standard deviation by n\sqrt{n} and using σ=20\sigma = 20 instead of σXˉ=5\sigma_{\bar{X}}=5. This gives Z=145−15020=−0.25Z = \frac{145-150}{20} = -0.25, and P(Z<−0.25)≈0.4013P(Z < -0.25) \approx 0.4013.
  • D: 0.8413 is P(Z>−1)P(Z > -1), which would answer the question 'what is the probability the mean weight is more than 145g?'.

Question 9

Let X1,X2,…,XnX_1, X_2, \dots, X_n be nn independent random variables, each with distribution N(μ,σ2)N(\mu, \sigma^2). Let S=∑i=1nXiS = \sum_{i=1}^n X_i be the sum and Xˉ=1nS\bar{X} = \frac{1}{n}S be the sample mean. Which of the following statements is incorrect?

  1. E(S)=nμE(S) = n\mu
  2. Var(S)=nσ2Var(S) = n\sigma^2
  3. Var(Xˉ)=σ2nVar(\bar{X}) = \frac{\sigma^2}{n}
  4. Var(S)=n2σ2Var(S) = n^2\sigma^2 (correct answer)
Explanation: Let's analyze each statement based on the properties of expectation and variance.\
  • A: E(S)=E(∑i=1nXi)=∑i=1nE(Xi)=∑i=1nμ=nμE(S) = E(\sum_{i=1}^n X_i) = \sum_{i=1}^n E(X_i) = \sum_{i=1}^n \mu = n\mu. This statement is correct.\
  • B: Var(S)=Var(∑i=1nXi)Var(S) = Var(\sum_{i=1}^n X_i). Because the variables are independent, Var(∑i=1nXi)=∑i=1nVar(Xi)=∑i=1nσ2=nσ2Var(\sum_{i=1}^n X_i) = \sum_{i=1}^n Var(X_i) = \sum_{i=1}^n \sigma^2 = n\sigma^2.** This statement is correct.\
  • C: Var(Xˉ)=Var(1nS)=(1n)2Var(S)=1n2(nσ2)=σ2nVar(\bar{X}) = Var(\frac{1}{n}S) = (\frac{1}{n})^2 Var(S) = \frac{1}{n^2} (n\sigma^2) = \frac{\sigma^2}{n}. This statement is correct.\
  • D: Var(S)=n2σ2Var(S) = n^2\sigma^2. This statement is incorrect. It confuses the variance of a sum of independent variables with the variance of a multiple of a single variable, i.e., Var(nX1)=n2Var(X1)=n2σ2Var(nX_1) = n^2 Var(X_1) = n^2\sigma^2. The correct formula for the sum is nσ2n\sigma^2 as shown in B.

Question 10

The number of calls received by a call centre in an hour follows a Poisson distribution with a mean of 40. Using a suitable approximation, find the probability that the call centre receives between 35 and 45 calls (inclusive) in a given hour.

  1. 0.109
  2. 0.523
  3. 0.571
  4. 0.615 (correct answer)
Explanation: Let XX be the number of calls. We are given X∼Po(40)X \sim Po(40). Since the mean λ=40\lambda = 40 is large (typically λ>10\lambda > 10), we can use a normal approximation Y∼N(λ,λ)Y \sim N(\lambda, \lambda), so Y∼N(40,40)Y \sim N(40, 40).
We want to find P(35≤X≤45)P(35 \le X \le 45). We must use a continuity correction.
P(35≤X≤45)P(35 \le X \le 45) is approximated by P(34.5<Y<45.5)P(34.5 < Y < 45.5).
The standard deviation is σ=40≈6.325\sigma = \sqrt{40} \approx 6.325.
We standardize the endpoints:
Z1=34.5−4040=−5.56.325≈−0.8696Z_1 = \frac{34.5 - 40}{\sqrt{40}} = \frac{-5.5}{6.325} \approx -0.8696
Z2=45.5−4040=5.56.325≈0.8696Z_2 = \frac{45.5 - 40}{\sqrt{40}} = \frac{5.5}{6.325} \approx 0.8696
We need P(−0.8696<Z<0.8696)=P(Z<0.8696)−P(Z<−0.8696)≈0.8077−0.1923=0.6154P(-0.8696 < Z < 0.8696) = P(Z < 0.8696) - P(Z < -0.8696) \approx 0.8077 - 0.1923 = 0.6154.\
  • A: 0.109 is the result of incorrectly using σ=40\sigma = 40 instead of σ=40\sigma = \sqrt{40}.
  • B: 0.523 is the result of using an incorrect continuity correction, such as P(35.5<Y<44.5)P(35.5 < Y < 44.5).
  • C: 0.571 is the result of not applying the continuity correction, calculating P(35<Y<45)P(35 < Y < 45).

Question 11

Let X1,X2,X3X_1, X_2, X_3 be independent random variables with Xi∼N(i,3)X_i \sim N(i, 3) for i=1,2,3i=1, 2, 3. Let Y=X1−2X2+X3Y = X_1 - 2X_2 + X_3. Find the variance of YY.

  1. -1
  2. 6
  3. 14
  4. 18 (correct answer)
Explanation: We are given that Xi∼N(i,3)X_i \sim N(i, 3) for i=1,2,3i=1, 2, 3, so each variable has variance 3.
X1∼N(1,3)X_1 \sim N(1, 3), X2∼N(2,3)X_2 \sim N(2, 3), X3∼N(3,3)X_3 \sim N(3, 3)
So Var(X1)=Var(X2)=Var(X3)=3Var(X_1) = Var(X_2) = Var(X_3) = 3.
The random variable YY is a linear combination Y=X1−2X2+X3Y = X_1 - 2X_2 + X_3.
For independent variables, the variance of a linear combination is:
Var(Y)=Var(X1−2X2+X3)=(1)2Var(X1)+(−2)2Var(X2)+(1)2Var(X3)Var(Y) = Var(X_1 - 2X_2 + X_3) = (1)^2Var(X_1) + (-2)^2Var(X_2) + (1)^2Var(X_3)
Var(Y)=1⋅3+4⋅3+1⋅3=3+12+3=18Var(Y) = 1 \cdot 3 + 4 \cdot 3 + 1 \cdot 3 = 3 + 12 + 3 = 18\
  • A: -1: Variance cannot be negative.
  • B: 6: This might result from incorrectly using the coefficients without squaring: (1+2+1)×3=12(1 + 2 + 1) \times 3 = 12 or from assuming variance 1 for each variable.
  • C: 14: This could come from summing 12+22+32=141^2 + 2^2 + 3^2 = 14 if the variances were i2i^2 and coefficients were all 1.

Question 12

A manufacturer claims that their light bulbs have a mean lifetime of 1000 hours. The lifetimes are not normally distributed, but have a known standard deviation of 120 hours. A consumer group tests a sample of 36 bulbs. Using the Central Limit Theorem, what is the probability that the sample mean lifetime is less than 950 hours?

  1. 0.0062 (correct answer)
  2. 0.0228
  3. 0.3385
  4. It cannot be determined without knowing the population distribution.
Explanation: We are given μ=1000\mu = 1000, σ=120\sigma = 120, and n=36n=36. The population is not normal.
However, since the sample size n=36n=36 is large enough (n≥30n \ge 30), we can apply the Central Limit Theorem. The CLT states that the sampling distribution of the sample mean Xˉ\bar{X} is approximately normal.
The mean of this distribution is μXˉ=μ=1000\mu_{\bar{X}} = \mu = 1000.
The standard deviation of this distribution (standard error) is σXˉ=σn=12036=1206=20\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{120}{\sqrt{36}} = \frac{120}{6} = 20.
So, Xˉ≈N(1000,202)\bar{X} \approx N(1000, 20^2).
We want to find P(Xˉ<950)P(\bar{X} < 950).
Standardize the value: Z=950−100020=−5020=−2.5Z = \frac{950 - 1000}{20} = \frac{-50}{20} = -2.5.
We need P(Z<−2.5)P(Z < -2.5). Using a calculator or standard normal table, this probability is approximately 0.0062.\
  • B: 0.0228 corresponds to a Z-score of -2. This would occur if the standard error was miscalculated as 25 (e.g. 1000−950=50,50/2=251000-950=50, 50/2 = 25).
  • C: 0.3385 is the result of forgetting to use the standard error and using the population standard deviation instead: Z=950−1000120≈−0.4167Z = \frac{950 - 1000}{120} \approx -0.4167, for which P(Z<−0.4167)≈0.3385P(Z < -0.4167) \approx 0.3385.
  • D: It cannot be determined... This is incorrect because the Central Limit Theorem allows us to approximate the probability due to the large sample size.

Question 13

Two independent random variables XX and YY are normally distributed, such that X∼N(100,122)X \sim N(100, 12^2) and Y∼N(95,52)Y \sim N(95, 5^2). Find the probability P(X>Y)P(X > Y), correct to three significant figures.

  1. 0.350
  2. 0.616
  3. 0.650 (correct answer)
  4. 0.677
Explanation: To find P(X>Y)P(X > Y), we consider the difference D=X−YD = X - Y. The distribution of DD is also normal.
E(D)=E(X)−E(Y)=100−95=5E(D) = E(X) - E(Y) = 100 - 95 = 5.\nSince XX and YY are independent, the variances add: Var(D)=Var(X)+Var(Y)=122+52=144+25=169Var(D) = Var(X) + Var(Y) = 12^2 + 5^2 = 144 + 25 = 169.\nThe standard deviation of DD is σD=169=13\sigma_D = \sqrt{169} = 13.
So, D∼N(5,132)D \sim N(5, 13^2).
The probability P(X>Y)P(X > Y) is equivalent to P(D>0)P(D > 0).
We standardize this value: Z=0−513=−513≈−0.3846Z = \frac{0 - 5}{13} = -\frac{5}{13} \approx -0.3846.
We need to find P(Z>−0.3846)P(Z > -0.3846). Using a calculator's normal cumulative distribution function, this is approximately 0.6498.
Rounded to three significant figures, the probability is 0.650.\
  • A: 0.350 is 1−0.6501 - 0.650, which corresponds to P(X<Y)P(X < Y).
  • B: 0.616 is the result of incorrectly adding standard deviations (σD=12+5=17\sigma_D = 12 + 5 = 17) instead of variances.
  • D: 0.677 is the result of incorrectly subtracting variances (Var(D)=144−25=119Var(D) = 144 - 25 = 119) because the variable is a difference.

Question 14

The time taken for a student to complete a puzzle is normally distributed with a mean of 15 minutes and a standard deviation of 4 minutes. A teacher takes a random sample of nn students. The probability that the mean time for the sample is less than 14 minutes is found to be 0.0228. Find the sample size nn.

  1. 4
  2. 8
  3. 16
  4. 64 (correct answer)
Explanation: Let XX be the time for a single student. X∼N(15,42)X \sim N(15, 4^2).
The sample mean Xˉ\bar{X} follows the distribution Xˉ∼N(15,42n)\bar{X} \sim N(15, \frac{4^2}{n}). The standard error is σXˉ=4n\sigma_{\bar{X}} = \frac{4}{\sqrt{n}}.
We are given P(Xˉ<14)=0.0228P(\bar{X} < 14) = 0.0228.
We first find the z-score corresponding to a cumulative probability of 0.0228. This is a standard value, z=−2z = -2.
Now we use the standardization formula: z=xˉ−μσXˉz = \frac{\bar{x} - \mu}{\sigma_{\bar{X}}}.
−2=14−154n-2 = \frac{14 - 15}{\frac{4}{\sqrt{n}}}.
−2=−14n-2 = \frac{-1}{\frac{4}{\sqrt{n}}}.
2=n42 = \frac{\sqrt{n}}{4}.
n=2×4=8\sqrt{n} = 2 \times 4 = 8.
n=82=64n = 8^2 = 64.\
  • A: 4 This can result from the algebraic error n4=12  ⟹  n=2  ⟹  n=4\frac{\sqrt{n}}{4} = \frac{1}{2} \implies \sqrt{n}=2 \implies n=4 or from 16n=z2=4  ⟹  n=4\frac{16}{n} = z^2 = 4 \implies n=4.
  • B: 8 This is the value of n\sqrt{n}, obtained by forgetting the final step of squaring.
  • C: 16 This would give a standard error of 4/16=14/\sqrt{16} = 1, leading to a Z-score of (14−15)/1=−1(14-15)/1 = -1, which corresponds to a probability of 0.1587, not 0.0228.

Question 15

Let X1,X2,…,XnX_1, X_2, \dots, X_n be nn independent random variables, each with distribution N(μ,σ2)N(\mu, \sigma^2). Let S=∑i=1nXiS = \sum_{i=1}^n X_i be the sum and Xˉ=1nS\bar{X} = \frac{1}{n}S be the sample mean. Which of the following statements is incorrect?

  1. E(S)=nμE(S) = n\mu
  2. Var(S)=nσ2Var(S) = n\sigma^2
  3. Var(Xˉ)=σ2nVar(\bar{X}) = \frac{\sigma^2}{n}
  4. Var(S)=n2σ2Var(S) = n^2\sigma^2 (correct answer)
Explanation: Let's analyze each statement based on the properties of expectation and variance.\
  • A: E(S)=E(∑i=1nXi)=∑i=1nE(Xi)=∑i=1nμ=nμE(S) = E(\sum_{i=1}^n X_i) = \sum_{i=1}^n E(X_i) = \sum_{i=1}^n \mu = n\mu. This statement is correct.\
  • B: Var(S)=Var(∑i=1nXi)Var(S) = Var(\sum_{i=1}^n X_i). Because the variables are independent, Var(∑i=1nXi)=∑i=1nVar(Xi)=∑i=1nσ2=nσ2Var(\sum_{i=1}^n X_i) = \sum_{i=1}^n Var(X_i) = \sum_{i=1}^n \sigma^2 = n\sigma^2.** This statement is correct.\
  • C: Var(Xˉ)=Var(1nS)=(1n)2Var(S)=1n2(nσ2)=σ2nVar(\bar{X}) = Var(\frac{1}{n}S) = (\frac{1}{n})^2 Var(S) = \frac{1}{n^2} (n\sigma^2) = \frac{\sigma^2}{n}. This statement is correct.\
  • D: Var(S)=n2σ2Var(S) = n^2\sigma^2. This statement is incorrect. It confuses the variance of a sum of independent variables with the variance of a multiple of a single variable, i.e., Var(nX1)=n2Var(X1)=n2σ2Var(nX_1) = n^2 Var(X_1) = n^2\sigma^2. The correct formula for the sum is nσ2n\sigma^2 as shown in B.

Question 16

Let X∼N(μ,σ2)X \sim N(\mu, \sigma^2). A random sample of size nn is taken. Consider the probabilities P1=P(X>μ+σ)P_1 = P(X > \mu + \sigma) and P2=P(Xˉ>μ+σ)P_2 = P(\bar{X} > \mu + \sigma), where Xˉ\bar{X} is the sample mean. For n>1n > 1, which of the following is true?

  1. P1=P2P_1 = P_2
  2. P1>P2P_1 > P_2 (correct answer)
  3. P1<P2P_1 < P_2
  4. The relationship depends on the value of nn.
Explanation: Let's analyze both probabilities by standardizing them.
For P1=P(X>μ+σ)P_1 = P(X > \mu + \sigma), the random variable is XX, which has mean μ\mu and standard deviation σ\sigma. The Z-score is Z=(μ+σ)−μσ=σσ=1Z = \frac{(\mu + \sigma) - \mu}{\sigma} = \frac{\sigma}{\sigma} = 1. So, P1=P(Z>1)P_1 = P(Z > 1).
For P2=P(Xˉ>μ+σ)P_2 = P(\bar{X} > \mu + \sigma), the random variable is Xˉ\bar{X}, which has mean μ\mu and standard deviation σXˉ=σn\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}. The Z-score is Z=(μ+σ)−μσ/n=σσ/n=nZ = \frac{(\mu + \sigma) - \mu}{\sigma/\sqrt{n}} = \frac{\sigma}{\sigma/\sqrt{n}} = \sqrt{n}. So, P2=P(Z>n)P_2 = P(Z > \sqrt{n}).
Since we are given n>1n > 1, we have n>1\sqrt{n} > 1.
The normal distribution's cumulative probability function is strictly increasing. Therefore, for z2>z1z_2 > z_1, P(Z>z2)<P(Z>z1)P(Z > z_2) < P(Z > z_1).
Since n>1\sqrt{n} > 1, it follows that P(Z>n)<P(Z>1)P(Z > \sqrt{n}) < P(Z > 1).
Therefore, P2<P1P_2 < P_1, which is the same as P1>P2P_1 > P_2.\
  • A: P1=P2P_1 = P_2 would only be true if n=1\sqrt{n}=1, i.e., n=1n=1, but the question specifies n>1n>1.
  • C: P1<P2P_1 < P_2 would imply n<1\sqrt{n}<1, which contradicts n>1n>1.
  • D: The relationship depends on the value of nn. The relationship P1>P2P_1 > P_2 holds for any n>1n>1, so it is not dependent on the specific value.

Question 17

The random variable XX is the sum of nn independent and identically distributed random variables, each with mean μ\mu and variance σ2\sigma^2. The random variable YY is the average of these nn variables. What are E(X)E(X) and Var(Y)Var(Y)?

  1. E(X)=nμE(X)=n\mu, Var(Y)=σ2/nVar(Y)=\sigma^2/n (correct answer)
  2. E(X)=nμE(X)=n\mu, Var(Y)=σ2/n2Var(Y)=\sigma^2/n^2
  3. E(X)=μE(X)=\mu, Var(Y)=σ2Var(Y)=\sigma^2
  4. E(X)=nμE(X)=n\mu, Var(Y)=nσ2Var(Y)=n\sigma^2
Explanation: Let the nn i.i.d. random variables be Z1,Z2,…,ZnZ_1, Z_2, \dots, Z_n, where E(Zi)=μE(Z_i)=\mu and Var(Zi)=σ2Var(Z_i)=\sigma^2.
The variable XX is the sum: X=∑i=1nZiX = \sum_{i=1}^n Z_i.
The expectation of the sum is the sum of the expectations: E(X)=E(∑Zi)=∑E(Zi)=∑μ=nμE(X) = E(\sum Z_i) = \sum E(Z_i) = \sum \mu = n\mu.
The variable YY is the average (sample mean): Y=Zˉ=1n∑Zi=1nXY = \bar{Z} = \frac{1}{n} \sum Z_i = \frac{1}{n}X.
The variance of the average is Var(Y)=Var(1nX)=(1n)2Var(X)Var(Y) = Var(\frac{1}{n}X) = (\frac{1}{n})^2 Var(X).
First, we need Var(X)Var(X). Since the variables are independent, the variance of the sum is the sum of the variances: Var(X)=Var(∑Zi)=∑Var(Zi)=∑σ2=nσ2Var(X) = Var(\sum Z_i) = \sum Var(Z_i) = \sum \sigma^2 = n\sigma^2.
Now, substitute this into the expression for Var(Y)Var(Y): Var(Y)=(1n)2(nσ2)=nσ2n2=σ2nVar(Y) = (\frac{1}{n})^2 (n\sigma^2) = \frac{n\sigma^2}{n^2} = \frac{\sigma^2}{n}.
Therefore, E(X)=nμE(X)=n\mu and Var(Y)=σ2/nVar(Y)=\sigma^2/n.\
  • B: Incorrectly calculates Var(Y)Var(Y) as σ2/n2\sigma^2/n^2, perhaps from squaring the 1/n1/n but forgetting the nn from Var(X)Var(X).
  • C: Incorrectly assumes the expectation of the sum is just μ\mu and the variance of the average is just σ2\sigma^2.
  • D: Correctly finds E(X)E(X) but incorrectly states that Var(Y)=nσ2Var(Y) = n\sigma^2. This is the variance of the sum XX, not the average YY.

Question 18

A machine produces bolts with lengths that are normally distributed with a standard deviation of 0.1 mm. The mean length, μ\mu, is adjustable. For a batch to be accepted, the probability that the mean length of a sample of 25 bolts is within 0.05 mm of μ\mu must be at least 0.99. What is the maximum possible value of the standard deviation of the population for this condition to be met?

  1. 0.0194 mm
  2. 0.0970 mm (correct answer)
  3. 0.4852 mm
  4. 2.5758 mm
Explanation: Let XX be the length of a bolt, X∼N(μ,σ2)X \sim N(\mu, \sigma^2). We are given n=25n=25. The sample mean Xˉ\bar{X} is distributed as Xˉ∼N(μ,σ2/25)\bar{X} \sim N(\mu, \sigma^2/25). The standard error is σXˉ=σ/25=σ/5\sigma_{\bar{X}} = \sigma/\sqrt{25} = \sigma/5.
We require the probability that the sample mean is 'within 0.05 mm of μ\mu' to be at least 0.99. This is written as P(∣Xˉ−μ∣≤0.05)≥0.99P(|\bar{X} - \mu| \le 0.05) \ge 0.99.
This is equivalent to P(−0.05≤Xˉ−μ≤0.05)≥0.99P(-0.05 \le \bar{X} - \mu \le 0.05) \ge 0.99.
To standardize, we divide by the standard error σ/5\sigma/5:
P(−0.05σ/5≤Z≤0.05σ/5)≥0.99P(\frac{-0.05}{\sigma/5} \le Z \le \frac{0.05}{\sigma/5}) \ge 0.99.
Let zc=0.05σ/5=0.25σz_c = \frac{0.05}{\sigma/5} = \frac{0.25}{\sigma}. We need P(−zc≤Z≤zc)≥0.99P(-z_c \le Z \le z_c) \ge 0.99.
This means the area in the two tails is at most 1−0.99=0.011 - 0.99 = 0.01. The area in the upper tail is at most 0.005. So P(Z>zc)≤0.005P(Z > z_c) \le 0.005.
This implies P(Z<zc)≥0.995P(Z < z_c) \ge 0.995.
Using the inverse normal function, the critical z-score for a cumulative probability of 0.995 is zcrit≈2.5758z_{crit} \approx 2.5758.
So we must have zc≥2.5758z_c \ge 2.5758.
0.25σ≥2.5758\frac{0.25}{\sigma} \ge 2.5758.
σ≤0.252.5758≈0.09705\sigma \le \frac{0.25}{2.5758} \approx 0.09705.
The maximum possible value for the standard deviation σ\sigma is approximately 0.0970 mm.\
  • A: 0.0194 is 0.05/2.57580.05 / 2.5758, from forgetting to use the standard error σ/5\sigma/5.
  • C: 0.4852 might come from an algebraic error, e.g., σ≤0.25×1.96\sigma \le 0.25 \times 1.96 using the wrong Z-score for 95% confidence.
  • D: 2.5758 is the Z-score itself, not the standard deviation.

Question 19

The time taken for a student to complete a puzzle is normally distributed with a mean of 15 minutes and a standard deviation of 4 minutes. A teacher takes a random sample of nn students. The probability that the mean time for the sample is less than 14 minutes is found to be 0.0228. Find the sample size nn.

  1. 4
  2. 8
  3. 16
  4. 64 (correct answer)
Explanation: Let XX be the time for a single student. X∼N(15,42)X \sim N(15, 4^2).
The sample mean Xˉ\bar{X} follows the distribution Xˉ∼N(15,42n)\bar{X} \sim N(15, \frac{4^2}{n}). The standard error is σXˉ=4n\sigma_{\bar{X}} = \frac{4}{\sqrt{n}}.
We are given P(Xˉ<14)=0.0228P(\bar{X} < 14) = 0.0228.
We first find the z-score corresponding to a cumulative probability of 0.0228. This is a standard value, z=−2z = -2.
Now we use the standardization formula: z=xˉ−μσXˉz = \frac{\bar{x} - \mu}{\sigma_{\bar{X}}}.
−2=14−154n-2 = \frac{14 - 15}{\frac{4}{\sqrt{n}}}.
−2=−14n-2 = \frac{-1}{\frac{4}{\sqrt{n}}}.
2=n42 = \frac{\sqrt{n}}{4}.
n=2×4=8\sqrt{n} = 2 \times 4 = 8.
n=82=64n = 8^2 = 64.\
  • A: 4 This can result from the algebraic error n4=12  ⟹  n=2  ⟹  n=4\frac{\sqrt{n}}{4} = \frac{1}{2} \implies \sqrt{n}=2 \implies n=4 or from 16n=z2=4  ⟹  n=4\frac{16}{n} = z^2 = 4 \implies n=4.
  • B: 8 This is the value of n\sqrt{n}, obtained by forgetting the final step of squaring.
  • C: 16 This would give a standard error of 4/16=14/\sqrt{16} = 1, leading to a Z-score of (14−15)/1=−1(14-15)/1 = -1, which corresponds to a probability of 0.1587, not 0.0228.

Question 20

Let X∼N(μ,σ2)X \sim N(\mu, \sigma^2). A random sample of size nn is taken. Consider the probabilities P1=P(X>μ+σ)P_1 = P(X > \mu + \sigma) and P2=P(Xˉ>μ+σ)P_2 = P(\bar{X} > \mu + \sigma), where Xˉ\bar{X} is the sample mean. For n>1n > 1, which of the following is true?

  1. P1=P2P_1 = P_2
  2. P1>P2P_1 > P_2 (correct answer)
  3. P1<P2P_1 < P_2
  4. The relationship depends on the value of nn.
Explanation: Let's analyze both probabilities by standardizing them.
For P1=P(X>μ+σ)P_1 = P(X > \mu + \sigma), the random variable is XX, which has mean μ\mu and standard deviation σ\sigma. The Z-score is Z=(μ+σ)−μσ=σσ=1Z = \frac{(\mu + \sigma) - \mu}{\sigma} = \frac{\sigma}{\sigma} = 1. So, P1=P(Z>1)P_1 = P(Z > 1).
For P2=P(Xˉ>μ+σ)P_2 = P(\bar{X} > \mu + \sigma), the random variable is Xˉ\bar{X}, which has mean μ\mu and standard deviation σXˉ=σn\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}. The Z-score is Z=(μ+σ)−μσ/n=σσ/n=nZ = \frac{(\mu + \sigma) - \mu}{\sigma/\sqrt{n}} = \frac{\sigma}{\sigma/\sqrt{n}} = \sqrt{n}. So, P2=P(Z>n)P_2 = P(Z > \sqrt{n}).
Since we are given n>1n > 1, we have n>1\sqrt{n} > 1.
The normal distribution's cumulative probability function is strictly increasing. Therefore, for z2>z1z_2 > z_1, P(Z>z2)<P(Z>z1)P(Z > z_2) < P(Z > z_1).
Since n>1\sqrt{n} > 1, it follows that P(Z>n)<P(Z>1)P(Z > \sqrt{n}) < P(Z > 1).
Therefore, P2<P1P_2 < P_1, which is the same as P1>P2P_1 > P_2.\
  • A: P1=P2P_1 = P_2 would only be true if n=1\sqrt{n}=1, i.e., n=1n=1, but the question specifies n>1n>1.
  • C: P1<P2P_1 < P_2 would imply n<1\sqrt{n}<1, which contradicts n>1n>1.
  • D: The relationship depends on the value of nn. The relationship P1>P2P_1 > P_2 holds for any n>1n>1, so it is not dependent on the specific value.