IB Mathematics: Analysis and Approaches Quiz: Normal Distribution

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Normal DistributionQuestion 1 of 20

The scores for two exams, Mathematics and Physics, are normally distributed. For Mathematics, scores $M$ follow $N(70, 12^2)$ . For Physics, scores $P$ follow $N(72, 8^2)$ . To receive a distinction, a student must score above 85. In which subject is it more likely for a student to receive a distinction?

Mathematics, because its standard deviation is larger.

Physics, because its corresponding z-score for a score of 85 is smaller.

Physics, because its mean is higher.

Mathematics, because its corresponding z-score for a score of 85 is smaller.

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Normal Distribution

Practice Normal Distribution in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Normal Distribution, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Mathematics, because its standard deviation is larger.
Physics, because its corresponding z-score for a score of 85 is smaller.
Physics, because its mean is higher.
Mathematics, because its corresponding z-score for a score of 85 is smaller. (correct answer)

Explanation: When comparing probabilities across different normal distributions, you need to standardize the values using z-scores to make meaningful comparisons. The key insight is that smaller z-scores (closer to the mean) correspond to higher probabilities in the tail beyond that point. To find which subject makes distinction more likely, calculate the z-score for a score of 85 in each subject. For Mathematics:

z_M = \frac{85 - 70}{12} = \frac{15}{12} = 1.25

. For Physics:

z_P = \frac{85 - 72}{8} = \frac{13}{8} = 1.625

. Since Mathematics has the smaller z-score (1.25 < 1.625), a score of 85 is closer to the mean in Mathematics, meaning more area lies above this point. Therefore, distinction is more likely in Mathematics. Option A incorrectly suggests that larger standard deviation alone makes distinction more likely, but you must consider how far 85 is from each mean relative to the spread. Option B states the opposite of what's true - Physics actually has the larger z-score, not smaller. Option C focuses only on the higher mean in Physics (72 vs 70) but ignores that Physics also has much smaller variability (σ = 8 vs 12), making scores above 85 relatively rarer despite the higher mean. Remember: when comparing probabilities across different normal distributions, always standardize using z-scores. The distribution with the smaller z-score for your threshold value will have the higher probability of exceeding that threshold.

Question 2

The heights of students in a school are normally distributed. Let $H$ be the height of a randomly chosen student. It is known that $P(H > 180) = 0.25$ and $P(H < 165) = 0.35$ . Find the conditional probability $P(H > 180 | H > 165)$ .

0.250
0.385 (correct answer)
0.400
0.615

Explanation: We need to find

P(H > 180 | H > 165)

. Using the formula for conditional probability,

P(A|B) = \frac{P(A \cap B)}{P(B)}

. Here, event A is

H > 180

and event B is

H > 165

. The intersection

A \cap B

is the event that

H > 180

AND

H > 165

, which simplifies to just

H > 180

. So, we need to calculate

\frac{P(H > 180)}{P(H > 165)}

. We are given

P(H > 180) = 0.25

. We are given

P(H < 165) = 0.35

, so

P(H > 165) = 1 - P(H < 165) = 1 - 0.35 = 0.65

. Therefore, the conditional probability is

\frac{0.25}{0.65} \approx 0.3846

Question 3

For a standard normal variable $Z \sim N(0, 1)$ , the statement $P(Z > c) = P(Z < -c)$ illustrates its symmetry. Which of the following statements correctly expresses this same property of symmetry for a general normal variable $X \sim N(\mu, \sigma^2)$ ?

$P(X > c) = P(X < -c)$ for any constant c.
$P(X > \mu) = P(X < \mu)$ for any constant c.
$P(X > c) = 1 - P(X < c)$ for any constant c.
$P(X > \mu + c) = P(X < \mu - c)$ for any constant c > 0. (correct answer)

Explanation: When you encounter questions about normal distribution symmetry, focus on how the center of symmetry changes from the standard normal to a general normal distribution. For the standard normal

Z \sim N(0, 1)

, the distribution is symmetric around 0, so

P(Z > c) = P(Z < -c)

means equal areas exist at equal distances from the center. For a general normal variable

X \sim N(\mu, \sigma^2)

, the center of symmetry shifts to the mean

\mu

, not zero. The correct symmetry property becomes:

P(X > \mu + c) = P(X < \mu - c)

for any constant

c > 0

. This states that the probability of being

c

units above the mean equals the probability of being

c

units below the mean, which perfectly captures the symmetric nature around

\mu

. This makes answer D correct. Let's examine why the other options fail: Option A incorrectly assumes the distribution is still symmetric around zero, ignoring that the mean has shifted to

\mu

. Option B has a logical error - it states "for any constant c" but doesn't actually involve

c

in the probability statement, and the statement

P(X > \mu) = P(X < \mu)

is always true regardless of

c

. Option C simply expresses the complement rule (

P(A) = 1 - P(A^c)

), which applies to any random variable and has nothing to do with symmetry. Study tip: For normal distribution symmetry questions, always identify the center first - it's 0 for standard normal, but

\mu

for general normal distributions. The symmetry property must reflect equal distances from this center.

Question 4

$P(X > c) = P(X < -c)$ for any constant c.
$P(X > \mu) = P(X < \mu)$ for any constant c.
$P(X > c) = 1 - P(X < c)$ for any constant c.
$P(X > \mu + c) = P(X < \mu - c)$ for any constant c > 0. (correct answer)

Z \sim N(0, 1)

, the distribution is symmetric around 0, so

P(Z > c) = P(Z < -c)

means equal areas exist at equal distances from the center. For a general normal variable

X \sim N(\mu, \sigma^2)

, the center of symmetry shifts to the mean

\mu

, not zero. The correct symmetry property becomes:

P(X > \mu + c) = P(X < \mu - c)

for any constant

c > 0

. This states that the probability of being

c

units above the mean equals the probability of being

c

units below the mean, which perfectly captures the symmetric nature around

\mu

. This makes answer D correct. Let's examine why the other options fail: Option A incorrectly assumes the distribution is still symmetric around zero, ignoring that the mean has shifted to

\mu

. Option B has a logical error - it states "for any constant c" but doesn't actually involve

c

in the probability statement, and the statement

P(X > \mu) = P(X < \mu)

is always true regardless of

c

. Option C simply expresses the complement rule (

P(A) = 1 - P(A^c)

\mu

for general normal distributions. The symmetry property must reflect equal distances from this center.

Question 5

The random variable $X$ follows a normal distribution $X \sim N(120, \sigma^2)$ . Given that $P(X < 135) = 0.92$ , find the standard deviation $\sigma$ .

10.6 (correct answer)
11.5
14.1
15.0

Explanation: First, we find the z-score corresponding to a cumulative probability of 0.92. Using an inverse normal function on a calculator,

z = \text{invNorm}(0.92, 0, 1) \approx 1.405

. The formula for the z-score is

z = \frac{x - \mu}{\sigma}

. Substitute the known values:

1.405 = \frac{135 - 120}{\sigma}

. This simplifies to

1.405 = \frac{15}{\sigma}

. Solving for

\sigma

gives

\sigma = \frac{15}{1.405} \approx 10.67

Question 6

The lifetimes of a brand of battery are normally distributed with a standard deviation of 30 hours. It is found that 18% of batteries have a lifetime of less than 450 hours. Find the mean lifetime, $\mu$ , of the batteries.

423 hours
477 hours (correct answer)
481 hours
504 hours

Explanation: We are given

P(X < 450) = 0.18

. First, find the z-score corresponding to this probability:

z = \text{invNorm}(0.18, 0, 1) \approx -0.915

. Using the standardization formula

z = \frac{x - \mu}{\sigma}

, we have

-0.915 = \frac{450 - \mu}{30}

. Multiply by 30:

-0.915 \times 30 = 450 - \mu

, which gives

-27.45 = 450 - \mu

. Solving for

\mu

\mu = 450 + 27.45 = 477.45

. Thus, the mean lifetime is approximately 477 hours.

Question 7

0.250
0.385 (correct answer)
0.400
0.615

Explanation: We need to find

P(H > 180 | H > 165)

. Using the formula for conditional probability,

P(A|B) = \frac{P(A \cap B)}{P(B)}

. Here, event A is

H > 180

and event B is

H > 165

. The intersection

A \cap B

is the event that

H > 180

AND

H > 165

, which simplifies to just

H > 180

. So, we need to calculate

\frac{P(H > 180)}{P(H > 165)}

. We are given

P(H > 180) = 0.25

. We are given

P(H < 165) = 0.35

, so

P(H > 165) = 1 - P(H < 165) = 1 - 0.35 = 0.65

. Therefore, the conditional probability is

\frac{0.25}{0.65} \approx 0.3846

Question 8

Let $X \sim N(\mu, \sigma^2)$ . Given that $P(X < \mu - a) = 0.3$ , find $P(\mu - a < X < \mu + a)$ .

0.3
0.4 (correct answer)
0.6
0.7

Explanation: The normal distribution is symmetric about the mean

\mu

. Given

P(X < \mu - a) = 0.3

. By symmetry, the probability in the upper tail is the same:

P(X > \mu + a) = P(X < \mu - a) = 0.3

. The total probability under the curve is 1. The probability we want is the area between the two tails. So,

P(\mu - a < X < \mu + a) = 1 - P(X < \mu - a) - P(X > \mu + a)

P(\mu - a < X < \mu + a) = 1 - 0.3 - 0.3 = 0.4

Question 9

The scores on a test, $S$ , are normally distributed with a mean of 75 and a standard deviation of 8. A student's score of 65 has a z-score of $z_1$ , and a score of 90 has a z-score of $z_2$ . Find the value of $z_2 - z_1$ .

1.250
1.875
3.125 (correct answer)
25.00

Explanation: First, calculate the z-score for each score using the formula

z = \frac{s - \mu}{\sigma}

. For a score of 65:

z_1 = \frac{65 - 75}{8} = \frac{-10}{8} = -1.25

. For a score of 90:

z_2 = \frac{90 - 75}{8} = \frac{15}{8} = 1.875

. The question asks for the value of

z_2 - z_1

z_2 - z_1 = 1.875 - (-1.25) = 1.875 + 1.25 = 3.125

Question 10

The random variable $X$ is normally distributed with a mean $\mu$ and standard deviation $\sigma$ . If $P(X < 88) = P(X > 112)$ , what is the value of $\mu$ ?

12
24
100 (correct answer)
200

Explanation: The condition

P(X < a) = P(X > b)

for a normal distribution implies that the values

a

and

b

are symmetric with respect to the mean

\mu

. The mean must therefore be the midpoint of

a

and

b

. In this case,

a = 88

and

b = 112

\mu = \frac{88 + 112}{2} = \frac{200}{2} = 100

Question 11

The random variable $X$ is normally distributed with mean 40. Given that $P(X > 50) = 0.1$ , find $P(30 < X < 50)$ .

0.4
Cannot be determined
0.9
0.8 (correct answer)

Explanation: When you encounter a normal distribution problem with symmetry properties, look for ways to use the bell curve's balanced nature to your advantage. Since

X

is normally distributed with mean 40, the distribution is perfectly symmetric around this center point. You're given that

P(X > 50) = 0.1

, which means the area under the curve to the right of 50 is 0.1. Here's the key insight: because 50 is exactly 10 units above the mean (40), and the normal distribution is symmetric, the probability

P(X < 30)

must also equal 0.1, since 30 is exactly 10 units below the mean. The areas in both tails are equal. Now you can find

P(30 < X < 50)

. Since the total probability under any distribution equals 1, and you know both tail areas:

P(30 < X < 50) = 1 - P(X < 30) - P(X > 50) = 1 - 0.1 - 0.1 = 0.8

Looking at the wrong answers: A) 0.4 incorrectly assumes this covers only half the central area. B) "Cannot be determined" ignores the powerful symmetry property that makes this solvable without knowing the standard deviation. C) 0.9 mistakenly subtracts only one tail probability instead of both. The answer is D) 0.8. Strategy tip: In normal distribution problems, always check if you can exploit symmetry around the mean. When given information about one tail, you often get the opposite tail "for free" due to the bell curve's balanced shape.

Question 12

Let $X_1 \sim N(100, 10^2)$ and $X_2 \sim N(100, 15^2)$ . Which statement correctly compares the probabilities of these two distributions?

$P(90 < X_1 < 110) < P(90 < X_2 < 110)$
$P(90 < X_1 < 110) = P(90 < X_2 < 110)$
$P(90 < X_1 < 110) > P(90 < X_2 < 110)$ (correct answer)
The relationship cannot be determined without knowing the exact probabilities.

Explanation: Both distributions have the same mean

\mu = 100

. The interval

(90, 110)

is symmetric about this mean.

X_1

has a smaller standard deviation (

\sigma_1 = 10

) than

X_2

(

\sigma_2 = 15

). A smaller standard deviation means the data is more concentrated around the mean. Therefore, the probability of finding a value within a certain range around the mean is higher for the distribution with the smaller standard deviation. Thus,

P(90 < X_1 < 110) > P(90 < X_2 < 110)

. We can also verify this by calculating z-scores. For

X_1

, the interval corresponds to

-1 < Z < 1

. For

X_2

, the interval corresponds to

-10/15 < Z < 10/15

, i.e.,

-0.67 < Z < 0.67

. The area from -1 to 1 is greater than the area from -0.67 to 0.67.

Question 13

423 hours
477 hours (correct answer)
481 hours
504 hours

Explanation: We are given

P(X < 450) = 0.18

. First, find the z-score corresponding to this probability:

z = \text{invNorm}(0.18, 0, 1) \approx -0.915

. Using the standardization formula

z = \frac{x - \mu}{\sigma}

, we have

-0.915 = \frac{450 - \mu}{30}

. Multiply by 30:

-0.915 \times 30 = 450 - \mu

, which gives

-27.45 = 450 - \mu

. Solving for

\mu

\mu = 450 + 27.45 = 477.45

. Thus, the mean lifetime is approximately 477 hours.

Question 14

1.250
1.875
3.125 (correct answer)
25.00

Explanation: First, calculate the z-score for each score using the formula

z = \frac{s - \mu}{\sigma}

. For a score of 65:

z_1 = \frac{65 - 75}{8} = \frac{-10}{8} = -1.25

. For a score of 90:

z_2 = \frac{90 - 75}{8} = \frac{15}{8} = 1.875

. The question asks for the value of

z_2 - z_1

z_2 - z_1 = 1.875 - (-1.25) = 1.875 + 1.25 = 3.125

Question 15

The weight of a product, $W$ , is normally distributed with a mean of 500g and a standard deviation of 10g. A product is rejected if its weight is less than 485g. If 10 products are chosen at random, what is the probability that exactly two are rejected?

0.0668
0.113 (correct answer)
0.150
0.225

Explanation: This is a two-step problem. First, find the probability (p) that a single product is rejected. Then, use the binomial distribution to find the probability of exactly two rejections in a sample of 10. Step 1: Find p.

p = P(W < 485)

. Standardize the value 485:

z = \frac{485 - 500}{10} = \frac{-15}{10} = -1.5

p = P(Z < -1.5) \approx 0.0668

. Step 2: Use the binomial distribution

X \sim B(n=10, p=0.0668)

. We want to find

P(X = 2)

P(X = 2) = \binom{10}{2} (0.0668)^2 (1 - 0.0668)^8

\binom{10}{2} = \frac{10 \times 9}{2} = 45

P(X = 2) = 45 \times (0.0668)^2 \times (0.9332)^8 \approx 45 \times 0.004462 \times 0.5768 \approx 0.1156

. The closest answer is 0.113.

Question 16

A machine fills bags of sugar. The weight $W$ of sugar in a bag is normally distributed with mean 505 g and standard deviation 3 g. A bag is selected at random. What is the probability that its weight is between 500 g and 510 g?

0.0478
0.4522
0.9044 (correct answer)
0.9522

Explanation: We need to find

P(500 < W < 510)

. We can use a calculator's normal cumulative distribution function (normalCdf). normalCdf(lower=500, upper=510,

\mu

=505,

\sigma

=3). Alternatively, we can use z-scores: For

W = 500

z_1 = \frac{500 - 505}{3} = -\frac{5}{3} \approx -1.667

. For

W = 510

z_2 = \frac{510 - 505}{3} = \frac{5}{3} \approx 1.667

. We need to find

P(-1.667 < Z < 1.667)

. This is

P(Z < 1.667) - P(Z < -1.667)

. Using a standard normal table or calculator:

P(Z < 1.667) \approx 0.9522

P(Z < -1.667) \approx 0.0478

. The probability is

0.9522 - 0.0478 = 0.9044

Question 17

Mathematics, because its standard deviation is larger.
Physics, because its corresponding z-score for a score of 85 is smaller.
Physics, because its mean is higher.
Mathematics, because its corresponding z-score for a score of 85 is smaller. (correct answer)

z_M = \frac{85 - 70}{12} = \frac{15}{12} = 1.25

. For Physics:

z_P = \frac{85 - 72}{8} = \frac{13}{8} = 1.625

Question 18

0.0478
0.4522
0.9044 (correct answer)
0.9522

Explanation: We need to find

P(500 < W < 510)

. We can use a calculator's normal cumulative distribution function (normalCdf). normalCdf(lower=500, upper=510,

\mu

=505,

\sigma

=3). Alternatively, we can use z-scores: For

W = 500

z_1 = \frac{500 - 505}{3} = -\frac{5}{3} \approx -1.667

. For

W = 510

z_2 = \frac{510 - 505}{3} = \frac{5}{3} \approx 1.667

. We need to find

P(-1.667 < Z < 1.667)

. This is

P(Z < 1.667) - P(Z < -1.667)

. Using a standard normal table or calculator:

P(Z < 1.667) \approx 0.9522

P(Z < -1.667) \approx 0.0478

. The probability is

0.9522 - 0.0478 = 0.9044

Question 19

0.0668
0.113 (correct answer)
0.150
0.225

p = P(W < 485)

. Standardize the value 485:

z = \frac{485 - 500}{10} = \frac{-15}{10} = -1.5

p = P(Z < -1.5) \approx 0.0668

. Step 2: Use the binomial distribution

X \sim B(n=10, p=0.0668)

. We want to find

P(X = 2)

P(X = 2) = \binom{10}{2} (0.0668)^2 (1 - 0.0668)^8

\binom{10}{2} = \frac{10 \times 9}{2} = 45

P(X = 2) = 45 \times (0.0668)^2 \times (0.9332)^8 \approx 45 \times 0.004462 \times 0.5768 \approx 0.1156

. The closest answer is 0.113.

Question 20

The random variable $X \sim N(\mu, \sigma^2)$ . The value $x_1 = 80$ has a z-score of $-0.5$ . The value $x_2 = 110$ has a z-score of $2.0$ . Find the mean, $\mu$ .

86 (correct answer)
90
95
98

Explanation: We can set up a system of two linear equations using the z-score formula

z = (x - \mu) / \sigma

$-0.5 = \frac{80 - \mu}{\sigma} \implies -0.5\sigma = 80 - \mu$
$2.0 = \frac{110 - \mu}{\sigma} \implies 2.0\sigma = 110 - \mu$

From equation (1), we can write

\mu = 80 + 0.5\sigma

. From equation (2), we can write

\mu = 110 - 2.0\sigma

. Set the expressions for

\mu

equal to each other:

80 + 0.5\sigma = 110 - 2.0\sigma

2.5\sigma = 30

\sigma = \frac{30}{2.5} = 12

. Now substitute the value of

\sigma

back into one of the equations for

\mu

\mu = 80 + 0.5(12) = 80 + 6 = 86