Business Statistics Quiz: Normal Distribution Applications
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Normal Distribution ApplicationsQuestion 1 of 17

A service center's response times follow a normal distribution with mean 18 minutes and standard deviation 4 minutes. The center promises that 90% of calls will be answered within a specified time limit. However, management discovers that during peak hours, while the mean remains 18 minutes, the distribution becomes right-skewed with the same 90th percentile as before. If the center wants to maintain its 90% promise during peak hours, what percentage of calls during peak hours will actually be answered within the original time limit?

Exactly 90%, since the 90th percentile remains unchanged by construction
Approximately 85%, due to the right skewness concentrating more probability in the tail
Approximately 94%, because right skewness shifts probability toward lower values
Approximately 88%, reflecting the interaction between skewness and the fixed percentile
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Business Statistics Quiz

Business Statistics Quiz: Normal Distribution Applications

Practice Normal Distribution Applications in Business Statistics 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 Applications, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Statistics.

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

A service center's response times follow a normal distribution with mean 18 minutes and standard deviation 4 minutes. The center promises that 90% of calls will be answered within a specified time limit. However, management discovers that during peak hours, while the mean remains 18 minutes, the distribution becomes right-skewed with the same 90th percentile as before. If the center wants to maintain its 90% promise during peak hours, what percentage of calls during peak hours will actually be answered within the original time limit?

  1. Exactly 90%, since the 90th percentile remains unchanged by construction (correct answer)
  2. Approximately 85%, due to the right skewness concentrating more probability in the tail
  3. Approximately 94%, because right skewness shifts probability toward lower values
  4. Approximately 88%, reflecting the interaction between skewness and the fixed percentile
Explanation: The key insight is in the problem statement: 'the distribution becomes right-skewed with the same 90th percentile as before.' This explicitly states that the 90th percentile value doesn't change between normal and peak hours. The original time limit was set at the 90th percentile of the normal distribution (18 + 4 × 1.28 = 23.12 minutes). During peak hours, by construction, this same value (23.12 minutes) remains the 90th percentile of the new skewed distribution. Therefore, exactly 90% of calls will still be answered within this time limit. The skewness changes the shape of the distribution but doesn't affect the percentile value that was preserved. Choices B, C, and D incorrectly assume the percentile changes when the problem explicitly states it remains the same.

Question 2

A bottling company's filling machine dispenses soda into cans with a fill volume that is normally distributed with a mean of 355 ml and a standard deviation of 4 ml. Cans filled with less than 350 ml are considered under-filled and are sold at a loss of $0.20 per can. In a production run of 10,000 cans, what is the expected total loss due to under-filled cans?

  1. $212.00 (correct answer)
  2. $456.00
  3. $106.00
  4. $21.20
Explanation: This is a multi-step problem. First, calculate the probability of a single can being under-filled. An under-filled can has a volume less than 350 ml. We need to find P(X < 350). Calculate the z-score: z = (350 - 355) / 4 = -5 / 4 = -1.25. From the standard normal table, P(Z < -1.25) ≈ 0.1056. Second, calculate the expected number of under-filled cans in a batch of 10,000: Expected number = 10,000 * 0.1056 = 1056 cans. Third, calculate the total expected loss: Total loss = 1056 cans * $0.20/can = $211.20. The closest answer is $212.00.

Question 3

A retail manager estimates that the weekly demand for a particular brand of coffee is normally distributed with a mean of 80 bags and a standard deviation of 15 bags. To minimize stockouts, the manager wants to ensure there is only a 5% chance of running out of stock in any given week. What is the minimum number of bags that should be on hand at the beginning of the week to achieve this service level?

  1. 95 bags
  2. 105 bags (correct answer)
  3. 110 bags
  4. 120 bags
Explanation: This is a service level problem. The manager wants the probability of demand being greater than the inventory level (Q) to be 5%. This is equivalent to saying the probability of demand being less than or equal to Q should be 95%. We need to find the 95th percentile of the demand distribution. First, find the z-score that corresponds to the 95th percentile. From a standard normal table, P(Z < 1.645) ≈ 0.95. The required inventory level Q is calculated as Q = μ + zσ. Plugging in the values: Q = 80 + 1.645 * 15 = 80 + 24.675 = 104.675. Since the number of bags must be an integer, the manager should stock 105 bags to ensure the stockout probability is not more than 5%.

Question 4

The number of daily customers at a specialty cafe is normally distributed with a mean of 150 and a standard deviation of 20. The cafe has the capacity to serve a maximum of 175 customers per day. What is the approximate probability that the cafe's capacity will be exceeded on at least one day in a 5-day work week? (Assume daily demand is independent).

  1. 0.106
  2. 0.445 (correct answer)
  3. 0.530
  4. 0.894
Explanation: This is a two-stage probability problem. First, find the probability that capacity is exceeded on any single day, let's call it p. This is P(X > 175). The z-score is z = (175 - 150) / 20 = 25 / 20 = 1.25. The probability is P(Z > 1.25) = 1 - P(Z < 1.25) = 1 - 0.8944 = 0.1056. So, p ≈ 0.1056. Second, we need to find the probability of this event happening at least once in 5 independent days (trials). This is a binomial probability scenario. P(at least one) = 1 - P(none). The probability of capacity NOT being exceeded on a given day is 1 - p = 1 - 0.1056 = 0.8944. The probability of capacity not being exceeded on any of the 5 days is (1 - p)^5 = (0.8944)^5 ≈ 0.5746. Therefore, the probability of exceeding capacity on at least one day is 1 - 0.5746 = 0.4254. The closest answer is 0.445.

Question 5

A tech company is forecasting demand for a new smartphone. The marketing department believes the standard deviation of the first-month demand will be 5,000 units. The company wants to ensure that the probability of demand being less than 30,000 units is only 10%. What should be the target mean for the first-month demand forecast?

  1. 23,600 units
  2. 36,400 units (correct answer)
  3. 38,225 units
  4. 30,000 units
Explanation: This is a reverse normal distribution problem. We are given a probability and need to find the mean (μ). We are given P(X < 30,000) = 0.10. We need to find the z-score that corresponds to the 10th percentile. From a standard normal table, the z-score for which P(Z < z) = 0.10 is approximately z = -1.28. The z-score formula is z = (X - μ) / σ. We can rearrange this to solve for μ: μ = X - zσ. Plugging in the values: μ = 30,000 - (-1.28 * 5,000) = 30,000 + 6,400 = 36,400 units. The company needs to aim for a mean demand of 36,400 units.

Question 6

The daily demand for a product is normally distributed with a mean of 500 units and a standard deviation of 50 units. The company decides to stock 575 units each day. If there is a stockout (demand > 575), the company incurs a fixed contractual penalty of $1,000, regardless of the size of the shortfall. What is the expected daily penalty cost?

  1. $66.80 (correct answer)
  2. $158.70
  3. $933.20
  4. $668.00
Explanation: The expected penalty cost is the probability of a stockout multiplied by the penalty amount. First, we need to calculate the probability of a stockout, which is P(Demand > 575). We calculate the z-score: z = (575 - 500) / 50 = 75 / 50 = 1.5. The probability of a stockout is P(Z > 1.5) = 1 - P(Z < 1.5). From a standard normal table, P(Z < 1.5) ≈ 0.9332. So, P(Stockout) = 1 - 0.9332 = 0.0668. The expected daily penalty cost is then: E(Cost) = P(Stockout) * Penalty = 0.0668 * $1,000 = $66.80. Distractor C represents the cost if the penalty was applied for not stocking out (0.9332 * $1000).

Question 7

The lifetime of a specific model of LED light bulb is normally distributed with a mean of 25,000 hours and a standard deviation of 2,000 hours. A hotel purchases a carton of 16 such bulbs. What is the probability that the average lifetime of the bulbs in the carton is more than 26,000 hours?

  1. 0.023 (correct answer)
  2. 0.159
  3. 0.309
  4. 0.617
Explanation: This problem requires using the Central Limit Theorem for the sampling distribution of the mean. The lifetime of a single bulb (X) is N(25000, 2000). We are interested in the average lifetime of a sample of n=16 bulbs (x̄). The distribution of x̄ is normal with mean μ_x̄ = μ = 25,000 and standard deviation σ_x̄ = σ / √n. The standard error is σ_x̄ = 2000 / √16 = 2000 / 4 = 500 hours. We need to find P(x̄ > 26,000). The z-score is z = (26,000 - 25,000) / 500 = 1000 / 500 = 2.0. The probability is P(Z > 2.0) = 1 - P(Z < 2.0) = 1 - 0.9772 = 0.0228. The closest answer is 0.023. A common error is to use the population standard deviation (2000), which would give z=0.5 and P(Z>0.5) = 0.309 (Distractor C).

Question 8

The demand for a product on any given day is normally distributed with a mean of 100 units and a standard deviation of 10 units. Assuming demands on different days are independent, what is the standard deviation of the total demand over a 4-day period?

  1. 10 units
  2. 40 units
  3. 20 units (correct answer)
  4. 100 units
Explanation: When you encounter questions about combining independent random variables, you need to understand how variances (not standard deviations) add together. This is a fundamental property in probability theory that applies when events are independent. Since daily demands are independent and normally distributed, the total demand over 4 days will also be normally distributed. The mean of the total demand is simply 4×100=4004 \times 100 = 400 units. However, standard deviations don't add directly - variances do. For independent random variables, the variance of their sum equals the sum of their individual variances. Each day has a variance of 102=10010^2 = 100 units². Over 4 days, the total variance is 4×100=4004 \times 100 = 400 units². The standard deviation is the square root of variance, so 400=20\sqrt{400} = 20 units. Choice A (10 units) incorrectly assumes the standard deviation stays the same regardless of the number of days. Choice B (40 units) makes the common mistake of adding standard deviations directly (4×10=404 \times 10 = 40), but this violates the mathematical properties of variance. Choice D (100 units) appears to confuse the daily mean with the combined standard deviation. Remember this key pattern: when combining independent random variables, always work with variances first (add them), then take the square root to get the standard deviation. The formula is σtotal=n×σindividual2\sigma_{total} = \sqrt{n \times \sigma_{individual}^2} where n is the number of independent periods. This prevents the common trap of incorrectly adding standard deviations directly.

Question 9

A manufacturer wants to produce pistons with a diameter of 80 mm. The process is centered correctly at 80 mm. To be considered high-quality (Six Sigma level), the defect rate must be no more than 3.4 parts per million. This corresponds to specification limits being placed at ±6 standard deviations from the mean. If the specification limits are 79.94 mm and 80.06 mm, what is the maximum allowable standard deviation (σ) to achieve this quality level?

  1. 0.005 mm
  2. 0.020 mm
  3. 0.010 mm (correct answer)
  4. 0.060 mm
Explanation: When you encounter Six Sigma quality control problems, you're dealing with the relationship between specification limits and process variation. Six Sigma means the specification limits are placed at ±6 standard deviations from the process mean, ensuring extremely low defect rates. Here, the process is centered at 80 mm with specification limits of 79.94 mm and 80.06 mm. Since the process is centered, both limits are equidistant from the mean. The distance from the mean to either specification limit is 80.06−80.00=0.0680.06 - 80.00 = 0.06 mm or 80.00−79.94=0.0680.00 - 79.94 = 0.06 mm. For Six Sigma quality, this distance must equal 6 standard deviations: 6σ=0.066σ = 0.06 mm. Solving for σ: σ=0.066=0.01σ = \frac{0.06}{6} = 0.01 mm. Looking at the wrong answers: A) 0.005 mm would place the specification limits at only ±3 standard deviations (6×0.005=0.036 × 0.005 = 0.03 mm), which is Three Sigma quality, not Six Sigma. B) 0.020 mm would require specification limits at ±0.12±0.12 mm from the mean, much wider than given. D) 0.060 mm would only allow ±1 standard deviation within the specification limits, representing very poor quality control. The correct answer is C) 0.010 mm. Study tip: In Six Sigma problems, always remember that "Six Sigma" literally means 6 standard deviations from the mean to the specification limit. Calculate the distance from mean to spec limit, then divide by 6 to find the maximum allowable standard deviation.

Question 10

A pharmaceutical company produces tablets with active ingredient content that is normally distributed with mean 250 mg and standard deviation 12 mg. Regulatory requirements state that tablets must contain between 225 mg and 275 mg to be acceptable. The company is considering two process improvements: Option A reduces the standard deviation to 8 mg but shifts the mean to 248 mg, while Option B keeps the standard deviation at 12 mg but shifts the mean to 253 mg. Which option results in a higher percentage of acceptable tablets?

  1. Option A, because reduced variability more than compensates for the slight mean shift (correct answer)
  2. Option B, because centering the mean optimally outweighs increased variability effects
  3. Both options yield approximately the same percentage of acceptable tablets
  4. Neither option improves upon the current process acceptance rate significantly
Explanation: Current process: X ~ N(250, 12²). P(225 < X < 275) requires z-scores: (225-250)/12 = -2.08 and (275-250)/12 = 2.08. P(-2.08 < Z < 2.08) ≈ 0.9812 - 0.0188 = 0.9624 or 96.24%. Option A: X ~ N(248, 8²). Z-scores: (225-248)/8 = -2.875 and (275-248)/8 = 3.375. P(-2.875 < Z < 3.375) ≈ 0.9998 - 0.002 = 0.9978 or 99.78%. Option B: X ~ N(253, 12²). Z-scores: (225-253)/12 = -2.33 and (275-253)/12 = 1.83. P(-2.33 < Z < 1.83) ≈ 0.9664 - 0.0099 = 0.9565 or 95.65%. Option A gives the highest acceptance rate due to reduced variability. Choice B incorrectly assumes mean centering dominates. Choice C ignores the significant difference. Choice D incorrectly suggests no improvement.

Question 11

A retail chain tracks customer satisfaction scores, which are normally distributed with mean 7.2 and standard deviation 1.4 on a 10-point scale. Corporate policy requires that stores with average satisfaction below the 25th percentile of this distribution undergo mandatory retraining. The chain operates 150 stores.

If satisfaction scores at different stores are independent and each store's average is based on a large sample (so the normal approximation applies), approximately how many stores would you expect to require retraining in any given evaluation period?

  1. Approximately 15 stores, since this represents exactly 10% of all stores
  2. Approximately 38 stores, based on the 25th percentile threshold definition (correct answer)
  3. Approximately 52 stores, accounting for natural variation in store performance
  4. Approximately 23 stores, considering the sampling variability of store averages
Explanation: By definition, the 25th percentile divides the distribution so that 25% of observations fall below it. Since each store's satisfaction score follows the same normal distribution independently, we expect 25% of the 150 stores to fall below the 25th percentile: 0.25 × 150 = 37.5 ≈ 38 stores. This is a direct application of the percentile definition. Choice A confuses 25th percentile with 10%. Choice C suggests some other proportion without justification. Choice D incorrectly attempts to adjust for sampling variability, but the problem states that store averages are based on large samples, making the normal approximation valid and the 25% proportion exact.

Question 12

A manufacturing process for a bearing has a mean diameter of 2.000 inches and a standard deviation of 0.004 inches. The specification limits are 1.990 to 2.010 inches. Due to a calibration error, the process mean shifts to 2.005 inches, while the standard deviation remains unchanged. What is the new percentage of bearings that will be outside the specification limits?

  1. 0.62%
  2. 3.24%
  3. 10.56%
  4. 10.70% (correct answer)
Explanation: This problem requires calculating defect rates after a process mean shift. The new mean is μ = 2.005 and σ = 0.004. The specification limits are LSL = 1.990 and USL = 2.010. We need to calculate P(X < 1.990) + P(X > 2.010) with the new mean. For the lower tail: z_LSL = (1.990 - 2.005) / 0.004 = -0.015 / 0.004 = -3.75. P(Z < -3.75) is very small, approximately 0.00009. For the upper tail: z_USL = (2.010 - 2.005) / 0.004 = 0.005 / 0.004 = 1.25. P(Z > 1.25) = 1 - P(Z < 1.25) = 1 - 0.8944 = 0.1056. The total defect rate is approximately 0.00009 + 0.1056 = 0.10569, or about 10.57%. The closest answer is 10.70%. Distractors might come from calculating the defect rate before the shift (P(|Z| > 2.5) = 1.24%) or only calculating one tail of the new distribution.

Question 13

Weekly sales for a premium electronics product are normally distributed with a mean of $10,000 and a standard deviation of $2,000. The company categorizes performance: sales in the top 15% are 'Excellent', and sales in the next 10% (i.e., between the 75th and 85th percentiles) are 'Good'. What is the approximate range of weekly sales that would be classified as 'Good'?

  1. $10,000 to $11,350
  2. $12,070 to $13,290
  3. $11,350 to $12,070 (correct answer)
  4. $8,650 to $9,320
Explanation: When you encounter questions about normal distributions with percentile rankings, you need to convert percentiles to z-scores, then transform those z-scores back to the original units using the given mean and standard deviation. To find the "Good" sales range (75th to 85th percentiles), you first need the corresponding z-scores. The 75th percentile corresponds to a z-score of approximately 0.675, and the 85th percentile corresponds to a z-score of approximately 1.04. You can find these values in a standard normal table or use the fact that these are commonly tested percentiles. Next, convert these z-scores back to dollar amounts using the formula: X=μ+z×σX = \mu + z \times \sigma, where μ=10,000\mu = 10,000 and σ=2,000\sigma = 2,000. For the 75th percentile: X=10,000+0.675×2,000=11,350X = 10,000 + 0.675 \times 2,000 = 11,350 For the 85th percentile: X=10,000+1.04×2,000=12,080≈12,070X = 10,000 + 1.04 \times 2,000 = 12,080 \approx 12,070 Therefore, the "Good" range is approximately $11,350 to $12,070. Choice A ($10,000 to 11,350)representstherangefromthemeantothe75thpercentile,missingtheupperbound.ChoiceB(11,350) represents the range from the mean to the 75th percentile, missing the upper bound. Choice B (12,070 to 13,290)showsthe"Excellent"category(top1513,290) shows the "Excellent" category (top 15%, or 85th percentile and above). Choice D (8,650 to $9,320) represents a lower percentile range, likely the 15th to 25th percentiles. Remember: always identify which percentiles you need first, find the corresponding z-scores, then convert back to the original units. Practice memorizing common percentile z-scores like 0.675 (75th) and 1.04 (85th) for faster calculation.

Question 14

A machine packages cereal into boxes. The weight of the cereal in a box is normally distributed with a mean of 500 grams and a standard deviation of 12 grams. A quality control inspector takes a random sample of 9 boxes. A batch is rejected if the sample mean weight is less than 495 grams. What is the probability that a batch will be rejected?

  1. 0.106 (correct answer)
  2. 0.338
  3. 0.417
  4. 0.000
Explanation: This question involves the sampling distribution of the mean. The distribution of individual boxes is N(500, 12). The distribution of the sample mean of n=9 boxes, according to the Central Limit Theorem, is also normal with mean μ_x̄ = μ = 500 and standard deviation σ_x̄ = σ / √n. The standard error of the mean is σ_x̄ = 12 / √9 = 12 / 3 = 4 grams. The batch is rejected if the sample mean is less than 495 grams. We need to calculate P(x̄ < 495). The z-score is z = (495 - 500) / 4 = -5 / 4 = -1.25. The probability is P(Z < -1.25) ≈ 0.1056. The most common error is to use the population standard deviation (σ=12) instead of the standard error (σ_x̄=4), which would lead to z = (495-500)/12 ≈ -0.417 and a probability of 0.338 (Distractor B).

Question 15

The daily demand for gasoline at a certain station is normally distributed with a mean (μ) and a standard deviation (σ). Given that on a particular day the demand was above the mean, what is the probability that the demand was also greater than μ + 1.5σ?

  1. 0.067
  2. 0.134 (correct answer)
  3. 0.500
  4. 0.933
Explanation: This is a conditional probability problem. Let A be the event that demand X > μ + 1.5σ, and B be the event that demand X > μ. We want to find P(A | B). The formula for conditional probability is P(A | B) = P(A and B) / P(B). The event (A and B) is 'X > μ + 1.5σ AND X > μ'. If X is greater than μ + 1.5σ, it is automatically greater than μ. So, the intersection (A and B) is just event A. The problem reduces to P(A) / P(B). We know that for a normal distribution, P(B) = P(X > μ) = 0.5. For P(A), we find P(X > μ + 1.5σ), which is equivalent to P(Z > 1.5). From the z-table, P(Z > 1.5) = 1 - P(Z < 1.5) = 1 - 0.9332 = 0.0668. Therefore, P(A | B) = 0.0668 / 0.5 = 0.1336. The closest answer is 0.134.

Question 16

A pharmaceutical company is developing a new pill. The weight of the active ingredient is a critical quality characteristic and is normally distributed. The company wants to set symmetric specification limits (μ ± k) such that 99% of the pills are within these limits. If the standard deviation of the process is 0.5 mg, what should be the total width of the specification range (i.e., the difference between the upper and lower limits)?

  1. 1.17 mg
  2. 1.96 mg
  3. 2.33 mg
  4. 2.58 mg (correct answer)
Explanation: To have 99% of the pills within symmetric limits, we must leave 0.5% (or 0.005) in each tail of the normal distribution. We need to find the z-score that corresponds to an upper tail probability of 0.005, which is the same as a cumulative probability of 1 - 0.005 = 0.995. From a standard normal table, z ≈ 2.576. The upper specification limit (USL) is μ + 2.576σ and the lower specification limit (LSL) is μ - 2.576σ. The total width of the specification range is USL - LSL = (μ + 2.576σ) - (μ - 2.576σ) = 2 * 2.576 * σ. With σ = 0.5 mg, the width is 2 * 2.576 * 0.5 = 2.576 mg. The closest answer is 2.58 mg. A common mistake is to find the z-score for a one-tailed 99% probability (z=2.33, Distractor C).

Question 17

A logistics company's delivery times are normally distributed with mean 2.8 days and standard deviation 0.6 days. The company guarantees delivery within 4 days or provides a full refund. Due to recent supply chain disruptions, the standard deviation has increased to 0.9 days while the mean remains at 2.8 days. To maintain the same refund rate as before the disruption, what should the new delivery guarantee be?

  1. Approximately 4.6 days, maintaining the same percentile in the new distribution (correct answer)
  2. Approximately 5.2 days, accounting for the proportional increase in variation
  3. Approximately 4.3 days, using a conservative adjustment for increased uncertainty
  4. Approximately 4.9 days, balancing customer expectations with operational reality
Explanation: Original setup: X ~ N(2.8, 0.6²). The 4-day guarantee corresponds to P(X > 4) = P(Z > (4-2.8)/0.6) = P(Z > 2.0) ≈ 0.0228. To maintain the same 2.28% refund rate with new distribution X ~ N(2.8, 0.9²), we need the 97.72nd percentile: 2.8 + 0.9 × 2.0 = 2.8 + 1.8 = 4.6 days. This ensures P(X > 4.6) ≈ 0.0228, matching the original refund rate. Choice B incorrectly applies proportional scaling (0.9/0.6 = 1.5, so 4 × 1.5 = 6, but this gives 5.2 after adjustment). Choice C uses an arbitrary conservative factor. Choice D doesn't follow any statistical principle for maintaining equivalent risk levels.