IB Mathematics: Applications and Interpretation Quiz: Normal Distribution
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Normal DistributionQuestion 1 of 20

The scores on a standardized university entrance exam are normally distributed with a mean of 500 and a standard deviation of 80.

To be eligible for a scholarship, a student must score in the top 8% of all test takers. What is the minimum score required to be eligible for the scholarship?

508
580
613
656
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Normal Distribution

Practice Normal Distribution in IB Mathematics: Applications and Interpretation 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: Applications and Interpretation.

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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

The scores on a standardized university entrance exam are normally distributed with a mean of 500 and a standard deviation of 80.

To be eligible for a scholarship, a student must score in the top 8% of all test takers. What is the minimum score required to be eligible for the scholarship?

  1. 508
  2. 580
  3. 613 (correct answer)
  4. 656
Explanation: To be in the top 8%, a student's score must be at the 92nd percentile. We need to find the score XX such that P(Score<X)=0.92P(\text{Score} < X) = 0.92. Using a calculator's inverse normal distribution function with area = 0.92, μ=500\mu = 500, and σ=80\sigma = 80, we get X≈612.5X \approx 612.5. The minimum score is 613 (rounded to the nearest integer).

Question 2

The speeds of cars on a highway are normally distributed. The mean speed is 90 km/h and the standard deviation is 10 km/h. The speed limit is 100 km/h.

If a police officer observes 500 cars, what is the expected number of cars that are travelling above the speed limit?

  1. 79 (correct answer)
  2. 84
  3. 159
  4. 421
Explanation: First, find the probability that a single car is travelling above the speed limit, P(S>100)P(S > 100). Given μ=90\mu = 90 and σ=10\sigma = 10, we calculate P(S>100)=normalcdf(100,∞,90,10)P(S > 100) = \text{normalcdf}(100, \infty, 90, 10). The z-score is z=(100−90)/10=1z = (100 - 90)/10 = 1. P(Z>1)≈0.158655P(Z > 1) \approx 0.158655. The expected number of speeding cars out of 500 is n×p=500×0.158655≈79.3275n \times p = 500 \times 0.158655 \approx 79.3275. The closest integer value is 79.

Question 3

A variable XX is normally distributed, X∼N(μ,52)X \sim N(\mu, 5^2). Given that P(X<22)=0.81P(X < 22) = 0.81.

Find the value of the mean μ\mu.

  1. 17.6 (correct answer)
  2. 18.2
  3. 26.4
  4. 30.8
Explanation: We are given σ=5\sigma = 5 and P(X<22)=0.81P(X < 22) = 0.81. We need to find μ\mu. First, find the z-score corresponding to a cumulative probability of 0.81: z=invNorm(0.81,0,1)≈0.8779z = \text{invNorm}(0.81, 0, 1) \approx 0.8779. Now use the z-score formula: z=x−μσz = \frac{x - \mu}{\sigma}. 0.8779=22−μ50.8779 = \frac{22 - \mu}{5}. Solving for μ\mu: 5×0.8779=22−μ⇒4.3895=22−μ⇒μ=22−4.3895≈17.615 \times 0.8779 = 22 - \mu \Rightarrow 4.3895 = 22 - \mu \Rightarrow \mu = 22 - 4.3895 \approx 17.61. The closest value is 17.6.

Question 4

The time, TT, in minutes, that a customer spends in a supermarket is normally distributed with a mean μ\mu and a standard deviation σ=12\sigma = 12. It is known that 20% of customers spend more than 85 minutes in the store.

Find the value of μ\mu, the mean time spent in the supermarket, to the nearest minute.

  1. 75 (correct answer)
  2. 84
  3. 95
  4. 102
Explanation: We are given P(T>85)=0.20P(T > 85) = 0.20, which means P(T<85)=0.80P(T < 85) = 0.80. We need to find μ\mu given σ=12\sigma = 12. First, find the z-score corresponding to a cumulative probability of 0.80: z=invNorm(0.80,0,1)≈0.8416z = \text{invNorm}(0.80, 0, 1) \approx 0.8416. Now use the z-score formula: z=x−μσz = \frac{x - \mu}{\sigma}. 0.8416=85−μ120.8416 = \frac{85 - \mu}{12}. Solving for μ\mu: 12×0.8416=85−μ⇒10.099=85−μ⇒μ=85−10.099≈74.912 \times 0.8416 = 85 - \mu \Rightarrow 10.099 = 85 - \mu \Rightarrow \mu = 85 - 10.099 \approx 74.9. To the nearest minute, μ=75\mu = 75.

Question 5

The journey time to work for an employee is normally distributed with a mean of 45 minutes. The probability that the journey takes more than 55 minutes is 0.12.

Due to the symmetry of the normal distribution, what is the probability that the journey takes less than 35 minutes?

  1. 0.12 (correct answer)
  2. 0.24
  3. 0.38
  4. 0.88
Explanation: The mean is μ=45\mu = 45 minutes. The value 55 minutes is 55−45=1055 - 45 = 10 minutes above the mean. The value 35 minutes is 45−35=1045 - 35 = 10 minutes below the mean. Because the normal distribution is symmetric about the mean, the probability of being more than kk units above the mean is equal to the probability of being more than kk units below the mean. Therefore, P(T>55)=P(T<35)P(T > 55) = P(T < 35). Since P(T>55)=0.12P(T > 55) = 0.12, it follows that P(T<35)=0.12P(T < 35) = 0.12.

Question 6

The error in a temperature reading from a particular sensor is normally distributed with a mean of 0°C and a standard deviation of 0.2°C.

What is the probability that the magnitude of the error is greater than 0.25°C?

  1. 0.106
  2. 0.211 (correct answer)
  3. 0.789
  4. 0.894
Explanation: The magnitude of the error being greater than 0.25°C means the error is either less than -0.25°C or greater than 0.25°C. Let EE be the error. We need to find P(E<−0.25)+P(E>0.25)P(E < -0.25) + P(E > 0.25). P(E>0.25)=normalcdf(0.25,∞,0,0.2)≈0.1056P(E > 0.25) = \text{normalcdf}(0.25, \infty, 0, 0.2) \approx 0.1056. Due to symmetry, P(E<−0.25)P(E < -0.25) is also 0.1056. The total probability is 0.1056+0.1056=0.21120.1056 + 0.1056 = 0.2112. Rounded to three decimal places, the probability is 0.211.

Question 7

The cholesterol levels of a population are normally distributed with a mean of 190 mg/dL and a standard deviation of 25 mg/dL.

What is the interquartile range (IQR) of the cholesterol levels, correct to one decimal place?

  1. 16.9 mg/dL
  2. 25.0 mg/dL
  3. 33.7 mg/dL (correct answer)
  4. 173.1 mg/dL
Explanation: The interquartile range (IQR) is the difference between the third quartile (Q3, 75th percentile) and the first quartile (Q1, 25th percentile). Using the inverse normal function on a GDC: Q1 = invNorm(0.25,190,25)≈173.137\text{invNorm}(0.25, 190, 25) \approx 173.137. Q3 = invNorm(0.75,190,25)≈206.863\text{invNorm}(0.75, 190, 25) \approx 206.863. IQR = Q3 - Q1 = 206.863−173.137≈33.726206.863 - 173.137 \approx 33.726. To one decimal place, the IQR is 33.7 mg/dL.

Question 8

A student takes two different tests. In Chemistry, her score is 88, where the class mean was 80 and the standard deviation was 5. In Physics, her score is 84, where the class mean was 75 and the standard deviation was 6.

Which statement correctly compares the student's relative performance on the two tests?

  1. She performed better in Chemistry because her raw score was higher.
  2. She performed better in Physics because her z-score was higher.
  3. She performed better in Chemistry because her z-score was higher. (correct answer)
  4. She performed equally well in both subjects because the difference from the mean was similar.
Explanation: To compare relative performance, we calculate the z-score for each test. A higher z-score indicates a better performance relative to the group. Chemistry z-score: zchem=88−805=85=1.6z_{chem} = \frac{88 - 80}{5} = \frac{8}{5} = 1.6. Physics z-score: zphys=84−756=96=1.5z_{phys} = \frac{84 - 75}{6} = \frac{9}{6} = 1.5. Since 1.6>1.51.6 > 1.5, her performance was relatively better in Chemistry.

Question 9

A machine produces bolts whose diameters are normally distributed with a mean of 10.00 mm and a standard deviation of 0.04 mm. A bolt is considered defective if its diameter is less than 9.90 mm or greater than 10.10 mm.

What is the probability that a randomly selected bolt is defective?

  1. 0.0062
  2. 0.0124 (correct answer)
  3. 0.4938
  4. 0.9876
Explanation: A bolt is defective if D<9.90D < 9.90 or D>10.10D > 10.10. We can calculate these probabilities separately and add them. P(D<9.90)=normalcdf(−∞,9.90,10.00,0.04)≈0.00621P(D < 9.90) = \text{normalcdf}(-\infty, 9.90, 10.00, 0.04) \approx 0.00621. P(D>10.10)=normalcdf(10.10,∞,10.00,0.04)≈0.00621P(D > 10.10) = \text{normalcdf}(10.10, \infty, 10.00, 0.04) \approx 0.00621. Due to symmetry, these probabilities are the same. The total probability of being defective is 0.00621+0.00621=0.012420.00621 + 0.00621 = 0.01242. Alternatively, we can find the probability of a non-defective bolt, P(9.90<D<10.10)≈0.9876P(9.90 < D < 10.10) \approx 0.9876, and subtract from 1: 1−0.9876=0.01241 - 0.9876 = 0.0124.

Question 10

The daily rainfall in a town, RR, is modelled by a normal distribution with a mean of 4.2 mm and a standard deviation of 1.5 mm.

On what percentage of days is the rainfall between 2.0 mm and 5.0 mm?

  1. 34.1%
  2. 65.6% (correct answer)
  3. 70.3%
  4. 95.4%
Explanation: We need to calculate P(2.0<R<5.0)P(2.0 < R < 5.0) for a normal distribution with μ=4.2\mu = 4.2 and σ=1.5\sigma = 1.5. Using a GDC's normal cumulative distribution function: normalcdf(2.0,5.0,4.2,1.5)≈0.6563\text{normalcdf}(2.0, 5.0, 4.2, 1.5) \approx 0.6563. Converting this probability to a percentage gives 0.6563×100%≈65.6%0.6563 \times 100\% \approx 65.6\%.

Question 11

The heights of adult sunflowers, HH, are normally distributed with a mean of 2.5 metres and a standard deviation of 0.3 metres.

A sunflower is chosen at random. Given that its height is greater than 2.3 metres, what is the probability that its height is less than 2.8 metres?

  1. 0.589
  2. 0.748
  3. 0.788 (correct answer)
  4. 0.841
Explanation: We need to find the conditional probability P(H<2.8∣H>2.3)P(H < 2.8 | H > 2.3). The formula is P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}. Here, A∩BA \cap B is the event 2.3<H<2.82.3 < H < 2.8. Using a GDC: P(2.3<H<2.8)=normalcdf(2.3,2.8,2.5,0.3)≈0.5889P(2.3 < H < 2.8) = \text{normalcdf}(2.3, 2.8, 2.5, 0.3) \approx 0.5889. P(H>2.3)=normalcdf(2.3,109,2.5,0.3)≈0.7475P(H > 2.3) = \text{normalcdf}(2.3, 10^9, 2.5, 0.3) \approx 0.7475. Therefore, the conditional probability is 0.58890.7475≈0.7878\frac{0.5889}{0.7475} \approx 0.7878, which rounds to 0.788.

Question 12

The weights of coffee bags filled by a machine are normally distributed with a mean weight of 500 g. It is found that 10% of the bags weigh more than 515 g.

What is the standard deviation of the weights of the coffee bags, correct to two decimal places?

  1. 1.28 g
  2. 9.45 g
  3. 11.70 g (correct answer)
  4. 15.00 g
Explanation: Let WW be the weight. We are given μ=500\mu = 500 and P(W>515)=0.10P(W > 515) = 0.10. This implies P(W<515)=0.90P(W < 515) = 0.90. We find the z-score for a cumulative probability of 0.90: z=invNorm(0.90,0,1)≈1.282z = \text{invNorm}(0.90, 0, 1) \approx 1.282. Using the z-score formula z=x−μσz = \frac{x - \mu}{\sigma}, we get 1.282=515−500σ1.282 = \frac{515 - 500}{\sigma}. This simplifies to 1.282=15σ1.282 = \frac{15}{\sigma}. Solving for σ\sigma: σ=151.282≈11.70\sigma = \frac{15}{1.282} \approx 11.70 g.

Question 13

The weights of coffee bags filled by a machine are normally distributed with a mean weight of 500 g. It is found that 10% of the bags weigh more than 515 g.

What is the standard deviation of the weights of the coffee bags, correct to two decimal places?

  1. 1.28 g
  2. 9.45 g
  3. 11.70 g (correct answer)
  4. 15.00 g
Explanation: Let WW be the weight. We are given μ=500\mu = 500 and P(W>515)=0.10P(W > 515) = 0.10. This implies P(W<515)=0.90P(W < 515) = 0.90. We find the z-score for a cumulative probability of 0.90: z=invNorm(0.90,0,1)≈1.282z = \text{invNorm}(0.90, 0, 1) \approx 1.282. Using the z-score formula z=x−μσz = \frac{x - \mu}{\sigma}, we get 1.282=515−500σ1.282 = \frac{515 - 500}{\sigma}. This simplifies to 1.282=15σ1.282 = \frac{15}{\sigma}. Solving for σ\sigma: σ=151.282≈11.70\sigma = \frac{15}{1.282} \approx 11.70 g.

Question 14

A student takes two different tests. In Chemistry, her score is 88, where the class mean was 80 and the standard deviation was 5. In Physics, her score is 84, where the class mean was 75 and the standard deviation was 6.

Which statement correctly compares the student's relative performance on the two tests?

  1. She performed better in Chemistry because her raw score was higher.
  2. She performed better in Physics because her z-score was higher.
  3. She performed better in Chemistry because her z-score was higher. (correct answer)
  4. She performed equally well in both subjects because the difference from the mean was similar.
Explanation: To compare relative performance, we calculate the z-score for each test. A higher z-score indicates a better performance relative to the group. Chemistry z-score: zchem=88−805=85=1.6z_{chem} = \frac{88 - 80}{5} = \frac{8}{5} = 1.6. Physics z-score: zphys=84−756=96=1.5z_{phys} = \frac{84 - 75}{6} = \frac{9}{6} = 1.5. Since 1.6>1.51.6 > 1.5, her performance was relatively better in Chemistry.

Question 15

The time, TT, in minutes, that a customer spends in a supermarket is normally distributed with a mean μ\mu and a standard deviation σ=12\sigma = 12. It is known that 20% of customers spend more than 85 minutes in the store.

Find the value of μ\mu, the mean time spent in the supermarket, to the nearest minute.

  1. 75 (correct answer)
  2. 84
  3. 95
  4. 102
Explanation: We are given P(T>85)=0.20P(T > 85) = 0.20, which means P(T<85)=0.80P(T < 85) = 0.80. We need to find μ\mu given σ=12\sigma = 12. First, find the z-score corresponding to a cumulative probability of 0.80: z=invNorm(0.80,0,1)≈0.8416z = \text{invNorm}(0.80, 0, 1) \approx 0.8416. Now use the z-score formula: z=x−μσz = \frac{x - \mu}{\sigma}. 0.8416=85−μ120.8416 = \frac{85 - \mu}{12}. Solving for μ\mu: 12×0.8416=85−μ⇒10.099=85−μ⇒μ=85−10.099≈74.912 \times 0.8416 = 85 - \mu \Rightarrow 10.099 = 85 - \mu \Rightarrow \mu = 85 - 10.099 \approx 74.9. To the nearest minute, μ=75\mu = 75.

Question 16

The journey time to work for an employee is normally distributed with a mean of 45 minutes. The probability that the journey takes more than 55 minutes is 0.12.

Due to the symmetry of the normal distribution, what is the probability that the journey takes less than 35 minutes?

  1. 0.12 (correct answer)
  2. 0.24
  3. 0.38
  4. 0.88
Explanation: The mean is μ=45\mu = 45 minutes. The value 55 minutes is 55−45=1055 - 45 = 10 minutes above the mean. The value 35 minutes is 45−35=1045 - 35 = 10 minutes below the mean. Because the normal distribution is symmetric about the mean, the probability of being more than kk units above the mean is equal to the probability of being more than kk units below the mean. Therefore, P(T>55)=P(T<35)P(T > 55) = P(T < 35). Since P(T>55)=0.12P(T > 55) = 0.12, it follows that P(T<35)=0.12P(T < 35) = 0.12.

Question 17

A variable XX is normally distributed, X∼N(μ,52)X \sim N(\mu, 5^2). Given that P(X<22)=0.81P(X < 22) = 0.81.

Find the value of the mean μ\mu.

  1. 17.6 (correct answer)
  2. 18.2
  3. 26.4
  4. 30.8
Explanation: We are given σ=5\sigma = 5 and P(X<22)=0.81P(X < 22) = 0.81. We need to find μ\mu. First, find the z-score corresponding to a cumulative probability of 0.81: z=invNorm(0.81,0,1)≈0.8779z = \text{invNorm}(0.81, 0, 1) \approx 0.8779. Now use the z-score formula: z=x−μσz = \frac{x - \mu}{\sigma}. 0.8779=22−μ50.8779 = \frac{22 - \mu}{5}. Solving for μ\mu: 5×0.8779=22−μ⇒4.3895=22−μ⇒μ=22−4.3895≈17.615 \times 0.8779 = 22 - \mu \Rightarrow 4.3895 = 22 - \mu \Rightarrow \mu = 22 - 4.3895 \approx 17.61. The closest value is 17.6.

Question 18

The weight XX of a suitcase is normally distributed with mean 18 kg and standard deviation 2 kg.

The airline charges extra for any suitcase weighing over 20.5 kg. What percentage of suitcases will be charged extra?

  1. 4.8%
  2. 10.6% (correct answer)
  3. 12.5%
  4. 89.4%
Explanation: We need to calculate the probability P(X>20.5)P(X > 20.5) for a normal distribution with μ=18\mu = 18 and σ=2\sigma = 2. Using a GDC: P(X>20.5)=normalcdf(20.5,∞,18,2)P(X > 20.5) = \text{normalcdf}(20.5, \infty, 18, 2). The z-score is z=(20.5−18)/2=2.5/2=1.25z = (20.5 - 18) / 2 = 2.5 / 2 = 1.25. P(Z>1.25)≈0.1056P(Z > 1.25) \approx 0.1056. To express this as a percentage, we multiply by 100: 0.1056×100%=10.6%0.1056 \times 100\% = 10.6\%.

Question 19

The lifespan of a certain brand of smartphone battery is normally distributed. The manufacturer warrants the batteries for 30 months. They found that 3% of the batteries fail before the warranty expires. The standard deviation of the battery lifespans is 4 months.

What is the mean lifespan of this brand of smartphone battery, correct to one decimal place?

  1. 22.5 months
  2. 37.5 months (correct answer)
  3. 38.8 months
  4. 42.0 months
Explanation: Let TT be the lifespan. We are given P(T<30)=0.03P(T < 30) = 0.03 and σ=4\sigma = 4. We need to find μ\mu. First, find the z-score corresponding to a cumulative probability of 0.03: z=invNorm(0.03,0,1)≈−1.881z = \text{invNorm}(0.03, 0, 1) \approx -1.881. Now use the z-score formula: z=x−μσz = \frac{x - \mu}{\sigma}. So, −1.881=30−μ4-1.881 = \frac{30 - \mu}{4}. Solving for μ\mu: μ=30−4(−1.881)=30+7.524=37.524\mu = 30 - 4(-1.881) = 30 + 7.524 = 37.524. To one decimal place, the mean lifespan is 37.5 months.

Question 20

The weights of a species of fish are normally distributed with a mean of 1200 g. The standard deviation is unknown. It is found that a fish weighing 1050 g has a z-score of -2.5.

What is the standard deviation of the weights of this species of fish?

  1. 60 g (correct answer)
  2. 150 g
  3. 375 g
  4. 420 g
Explanation: The z-score formula is z=x−μσz = \frac{x - \mu}{\sigma}. We are given z=−2.5z = -2.5, x=1050x = 1050, and μ=1200\mu = 1200. We need to find σ\sigma. Substitute the given values into the formula: −2.5=1050−1200σ-2.5 = \frac{1050 - 1200}{\sigma}. −2.5=−150σ-2.5 = \frac{-150}{\sigma}. Solving for σ\sigma: σ=−150−2.5=60\sigma = \frac{-150}{-2.5} = 60. The standard deviation is 60 g.