A pharmacy is setting a service target for prescription fill times, which are approximately normal with standard deviation σ = 4 minutes. The pharmacy wants about 4% of prescriptions to take longer than 20 minutes. What mean fill time does it need to hit?

z = 1.75 for the 96th percentile, so μ = 20 - (1.75)(4) = 13 minutes.

Normal distribution practice problems and solutions

Q: Can I compare a value from one normal distribution with a value from another?

Yes, and that is exactly what a z-score is for. Standardize each value inside its own distribution, then compare the z-scores; the raw values themselves are not comparable when the means, the standard deviations, or the units differ. The same idea runs one level up as well: two whole processes can be compared by the area each one puts past a shared boundary rather than by their means, which is how a process with the lower mean can still be the safer one. One caution on direction: when a smaller number is the better outcome, such as a wait time, the more negative z is the better result, so read the context before deciding which score wins.

By Jude Wallis · Published July 29, 2026

This set has 8 applied normal distribution problems. Most run forward: name the variable and its distribution, standardize the boundary, read the area, answer in context. Others run backward, from a percent to a value or to the mean or standard deviation itself. Solve each on paper first.

AP Statistics: Unit 2 (topics 2.11 The Normal Distribution). In the Fall 2026 AP Statistics course, finding areas and percentiles under a normal curve, working backward from a percentile to a value, and using a known percentile to recover a parameter all sit in Unit 2, topic 2.11 (The Normal Distribution), and the exam supplies the standard normal table these solutions read.

What these problems build

These 8 problems build one habit: every normal-curve question is a trip between a raw value and an area, and the only decision is which direction you are traveling. Going forward, you start with a value $x$ , standardize it with $z = \frac{x - \mu}{\sigma}$ , and read an area. Going backward, you start with a percent, find the $z$ that produces that area, and unstandardize with $x = \mu + z\sigma$ . Two problems here run backward one step further, past the value to a parameter: given a boundary and the percent that has to sit beyond it, they solve for the mean or the standard deviation a process would need. Here $\mu$ (mu) is the population mean and $\sigma$ (sigma) is the population standard deviation, so these are parameters and the $z$ -score is exact, not an estimate.

The four steps that earn credit on an AP free-response question are the same every time:

Define the variable and state the distribution, for example: let $S$ be the height of a randomly chosen seedling, in centimeters, with $S$ approximately $N(\mu = 12, \sigma = 2.5)$ . Then sketch the curve with the region shaded.
Standardize the boundary, or work backwards from the area if the question hands you a percentile.
Do the table work: read the area, subtract areas, or search the body of the table for the $z$ you need.
Answer in context, with a percent or a value carrying its units.

Skipping step 1 or step 4 is where most of the lost points live. A bare decimal sitting alone under a question about seedling heights is not an answer to that question.

The five question shapes in this set

Every problem below is one of five shapes. Name the shape before you touch the arithmetic, because the shape tells you which way the calculation runs.

Question sounds like	Direction	What you do
What percent are above (or below) a value?	Value to area	Standardize, read the table, subtract from 1 if you need the right tail
What percent fall between two limits, or outside them?	Value to area	Standardize both, subtract the smaller left area from the larger; for outside, add the two tails
What value marks the top 10%, or the middle 60%?	Area to value	Find $z$ from the area, then $x = \mu + z\sigma$
What mean or standard deviation would hit a target percent?	Area to parameter	Find $z$ from the area, then solve $x = \mu + z\sigma$ for the unknown letter
Which of two results, or two processes, is better?	Value to $z$ , or area to area	Standardize inside each distribution and compare

One move stacks on top of the first shape. Once you have a proportion, an expected count is that proportion times the number of trials, and an expected count does not have to come out a whole number.

The empirical rule is the shortcut for the special case where every boundary lands exactly 1, 2, or 3 standard deviations from the mean. Almost every boundary in this set lands somewhere else, which is why the table does the work here. The rule earns its keep as a sanity check instead: any central band narrower than 68% has to sit inside the one-standard-deviation range, and a band that comes out wider than that is an arithmetic slip.

How to read the table, and how close is close enough

The z-table gives the area to the left of a $z$ -score, and it indexes $z$ to two decimal places. Two consequences run through every solution here.

First, a right tail is always $1 - (\text{area to the left})$ , and an area between two values is always (larger left area) minus (smaller left area). Adding two left areas is the single most common wrong move on this material.

Second, going backwards means searching the body of the table for an area and reading off the $z$ at its edge, so you almost never land on the target exactly. Take the nearest entry: for an area of 0.9000 the table offers 0.8997 at $z = 1.28$ and 0.9015 at $z = 1.29$ , so you use $z = 1.28$ . Because $z$ is rounded to two decimals, a table answer and a calculator answer differ in the third or fourth decimal place, and the solutions below flag that gap wherever it pushes an answer past a limit the question set. Both are correct on the exam; showing the standardized boundary is what gets scored.

Work each problem with pencil and paper first. To review the standardizing step, see how to find a z-score; to check a single area, open the normal distribution calculator or the empirical rule calculator; for the z-score fundamentals behind this set, try z-scores and normal distribution practice or the topic page for 2.11 The Normal Distribution.

Frequently asked questions

Does the z-table give the area to the left or to the right?

The standard normal table used on the AP exam gives the area to the left of a $z$ -score, which is the same as the percentile of that value. A right tail is $1 - (\text{area to the left})$ , so a boundary carrying 0.8770 to its left leaves $1 - 0.8770 = 0.1230$ above it. An area between two values is the larger left area minus the smaller left area, never the two left areas added, which would double-count everything below the lower boundary. If your answer for a right tail comes out above 0.5 for a value above the mean, you almost certainly forgot to subtract.

When can I use the empirical rule instead of the table?

Only when every boundary in the question lands exactly 1, 2, or 3 standard deviations from the mean. That is a narrow case, and this set is built to sit outside it: a boundary at $z = 1.4$ or $z = -0.6$ leaves the 68-95-99.7 percentages with nothing to say, and the table has to do the work. Even a boundary that does land on a landmark is served better by the table, whose 0.0013 below $z = -3$ beats the rule's rounded 0.15%. Treat the rule as a sanity check on a table answer instead: a central band narrower than 68% has to sit inside the one-standard-deviation range, and a band wider than that means an arithmetic slip.

How do I go from a percentile back to a value?

Reverse the order of operations. Convert the wording into an area to the left, search the body of the table for that area, read the $z$ at its edge, then unstandardize with $x = \mu + z\sigma$ . Watch the translation step: the top 10% means an area of 0.90 to the left, not 0.10, and a middle 90% means 0.05 in each tail, so the two boundaries carry 0.05 and 0.95 to their left. The same reversal recovers a parameter: when the value is known and $\mu$ or $\sigma$ is not, put the numbers you have into $x = \mu + z\sigma$ and solve for the letter you want instead of for $x$ .

Can I compare a value from one normal distribution with a value from another?

Yes, and that is exactly what a $z$ -score is for. Standardize each value inside its own distribution, then compare the $z$ -scores; the raw values themselves are not comparable when the means, the standard deviations, or the units differ. The same idea runs one level up as well: two whole processes can be compared by the area each one puts past a shared boundary rather than by their means, which is how a process with the lower mean can still be the safer one. One caution on direction: when a smaller number is the better outcome, such as a wait time, the more negative $z$ is the better result, so read the context before deciding which score wins.

Problem 1

A city bike-share program finds that trip durations are approximately normal with mean $\mu = 22$ minutes and standard deviation $\sigma = 5$ minutes. What proportion of trips last longer than 30 minutes?

Show the worked solution

State the distribution. Let $X$ be the duration of a randomly chosen trip, in minutes. $X$ is approximately $N(\mu = 22, \sigma = 5)$ , and the question asks for $P(X > 30)$ . Sketch the curve and shade to the right of 30.
Standardize the boundary. $z = \frac{x - \mu}{\sigma} = \frac{30 - 22}{5} = \frac{8}{5} = 1.60$ , so 30 minutes sits 1.60 standard deviations above the mean.
Read the area. The table gives area to the left, and $z = 1.60$ has 0.9452 to its left. The shaded right tail is $1 - 0.9452 = 0.0548$ .
Answer in context. About 5.48% of bike-share trips last longer than 30 minutes.

$z = 1.60$ and the area to the right is $1 - 0.9452 = 0.0548$ , so about 5.48% of trips run longer than 30 minutes.

Problem 2

A greenhouse grows a tomato variety whose fruit weights are approximately normal with mean $\mu = 180$ g and standard deviation $\sigma = 24$ g. A grader routes fruit to the premium bin when it weighs between 162 g and 210 g. What percent of the crop goes to the premium bin?

Show the worked solution

State the distribution. Let $X$ be the weight of a randomly chosen tomato, in grams, with $X$ approximately $N(\mu = 180, \sigma = 24)$ . The question asks for $P(162 < X < 210)$ , the shaded strip between the two boundaries.
Standardize both boundaries. Lower: $z = \frac{162 - 180}{24} = \frac{-18}{24} = -0.75$ . Upper: $z = \frac{210 - 180}{24} = \frac{30}{24} = 1.25$ .
Read both left areas and subtract. The table gives 0.2266 to the left of $z = -0.75$ and 0.8944 to the left of $z = 1.25$ , so the area between is $0.8944 - 0.2266 = 0.6678$ . Subtract, never add: adding the two left areas would double-count everything below 162 g.
Answer in context. About 66.78% of the crop weighs between 162 g and 210 g and goes to the premium bin.

$z = -0.75$ and $z = 1.25$ , and $0.8944 - 0.2266 = 0.6678$ , so about 66.78% of the crop goes to the premium bin.

Problem 3

A tablet press is set to produce 250 mg tablets, and the plant rejects any tablet outside 240 mg to 260 mg. This week the press is running with mean $\mu = 251$ mg and standard deviation $\sigma = 4$ mg, and tablet mass is approximately normal. (a) What percent of tablets fall outside the accepted range? (b) Which side of the range loses more tablets, and what does that say about where the press is centered?

Show the worked solution

State the distribution and the two boundaries. Let $X$ be the mass of a randomly chosen tablet, in milligrams, approximately $N(\mu = 251, \sigma = 4)$ . Out of spec means $X < 240$ or $X > 260$ , so the shaded region is the pair of tails outside the accepted range, not the middle.
(a) Standardize both limits. Lower: $z = \frac{240 - 251}{4} = \frac{-11}{4} = -2.75$ . Upper: $z = \frac{260 - 251}{4} = \frac{9}{4} = 2.25$ . The mean sits 1 mg above the 250 mg target, so the two limits are not the same distance away in standard deviations.
(a) Read each tail. The table gives 0.0030 to the left of $z = -2.75$ , which is the low tail directly. It gives 0.9878 to the left of $z = 2.25$ , so the high tail is $1 - 0.9878 = 0.0122$ .
(a) Add the two tails. $0.0030 + 0.0122 = 0.0152$ , about 1.52% of tablets. Check it the other way as well: the area between the limits is $0.9878 - 0.0030 = 0.9848$ , and $1 - 0.9848 = 0.0152$ , the same answer.
(b) Compare the two tails. The high side loses 1.22% and the low side loses 0.30%, so roughly four times as many tablets are rejected for being heavy as for being light. A lopsided pair of tails is the signature of a mean that has drifted off target, and here it has drifted up. Recentring the press at 250 mg would put both limits 2.5 standard deviations out, cutting the total loss to $2(0.0062) = 0.0124$ , about 1.24%.
Answer in context. About 1.52% of tablets fall outside the 240 mg to 260 mg range, and most of that loss is on the heavy side because the press is running 1 mg above its 250 mg target.

(a) $z = -2.75$ and $z = 2.25$ , so $0.0030 + 0.0122 = 0.0152$ , about 1.52% out of spec; (b) the high side, at 1.22% against 0.30%, because the press is centered 1 mg above the 250 mg target.

Problem 4

A pharmacy is setting a service target for prescription fill times, which are approximately normal with standard deviation $\sigma = 4$ minutes. The pharmacy wants about 4% of prescriptions to take longer than 20 minutes. What mean fill time does it need to hit?

Show the worked solution

State what is known and what is missing. Let $X$ be the fill time of a randomly chosen prescription, in minutes, approximately $N(\mu, \sigma = 4)$ with $\mu$ unknown. The requirement $P(X > 20) = 0.04$ means 20 minutes has to sit at the 96th percentile, with $1 - 0.04 = 0.96$ of the area to its left. This runs backward, and one step further than usual: the unknown is a parameter, not a value.
Find the z-score from the area. Search the body of the table for 0.9600. The nearest entries are 0.9599 at $z = 1.75$ and 0.9608 at $z = 1.76$ , and 0.9599 is closer, so $z = 1.75$ . The 20-minute mark has to sit 1.75 standard deviations above the mean.
Solve for the mean. Rearranging $x = \mu + z\sigma$ gives $\mu = x - z\sigma = 20 - (1.75)(4)$ . Multiply first: $(1.75)(4) = 7$ , so $\mu = 20 - 7 = 13$ minutes.
Check the answer and the rounding direction. With $\mu = 13$ the boundary standardizes back to $z = \frac{20 - 13}{4} = 1.75$ , so the check closes. The table's 1.75 is a shade short of the exact 1.7507, which leaves 4.01% above 20 minutes rather than 4.00%; unrounded technology puts the target at 12.997 minutes, the same 13.0 to a tenth of a minute.
Answer in context. The pharmacy needs to run to a mean fill time of about 13 minutes. Note which lever it pulled: with the spread fixed at 4 minutes, the only way to meet the target is to move the mean. If the pharmacy could instead tighten the spread to $\sigma = 3$ minutes, a mean of $20 - (1.75)(3) = 14.75$ minutes would meet the same 4% target.

$z = 1.75$ for the 96th percentile, so $\mu = 20 - (1.75)(4) = 13$ minutes.

Problem 5

Two machines fill bottles labeled 500 mL, and a bottle counts as short if it holds less than 500 mL. Machine A fills to a volume that is approximately normal with $\mu = 502$ mL and $\sigma = 0.8$ mL. Machine B fills to a volume that is approximately normal with $\mu = 503$ mL and $\sigma = 1.5$ mL. (a) What percent of bottles does each machine send out short? (b) Machine B pours more on average, so explain why it still sends out more short bottles.

Show the worked solution

State both distributions and the shared boundary. Let $A$ be the volume from machine A and $B$ the volume from machine B, in milliliters, with $A$ approximately $N(\mu = 502, \sigma = 0.8)$ and $B$ approximately $N(\mu = 503, \sigma = 1.5)$ . Both parts turn on the left tail below 500 mL, and each one has to be found inside its own machine's distribution.
(a) Standardize 500 mL for machine A. $z = \frac{500 - 502}{0.8} = \frac{-2}{0.8} = -2.50$ . The table gives 0.0062 to the left, so about 0.62% of machine A's bottles are short.
(a) Standardize 500 mL for machine B. $z = \frac{500 - 503}{1.5} = \frac{-3}{1.5} = -2.00$ . The table gives 0.0228 to the left, so about 2.28% of machine B's bottles are short. This boundary happens to land exactly 2 standard deviations out, where the empirical rule would round to 2.5%; with a table in hand, report the table's 0.0228.
(b) Compare the two cushions in standard deviations, not in milliliters. Machine B's mean sits 3 mL above the label against machine A's 2 mL, but machine B's spread is nearly twice as wide. Those 3 mL buy only $\frac{3}{1.5} = 2.00$ standard deviations of cushion, while machine A's 2 mL buy $\frac{2}{0.8} = 2.50$ . A $z$ -score measures distance from the mean in standard deviations, so the tighter machine sits further from the label even though it pours less.
Answer in context. About 0.62% of machine A's bottles and 2.28% of machine B's bottles fall below the 500 mL label, so machine A is the safer machine to run. A higher mean does not protect a label on its own; extra spread hands part of the curve straight back across the boundary.

(a) Machine A: $z = -2.50$ , about 0.62% short. Machine B: $z = -2.00$ , about 2.28% short. (b) Machine B's 3 mL cushion is only 2.00 standard deviations wide while machine A's 2 mL cushion is 2.50, so the tighter machine wins.

Problem 6

A laptop maker reports that battery life on one model is approximately normal with mean $\mu = 11.2$ hours and standard deviation $\sigma = 0.8$ hours. The warranty replaces any unit that runs under 10 hours. (a) What proportion of units qualify for replacement? (b) In a shipment of 500 units, how many should the maker expect to qualify?

Show the worked solution

State the distribution. Let $X$ be the battery life of a randomly chosen unit, in hours, approximately $N(\mu = 11.2, \sigma = 0.8)$ . Part (a) asks for $P(X < 10)$ , the left tail below 10 hours.
(a) Standardize the boundary. $z = \frac{10 - 11.2}{0.8} = \frac{-1.2}{0.8} = -1.50$ , so the warranty threshold sits 1.50 standard deviations below the mean.
(a) Read the area. The table gives 0.0668 to the left of $z = -1.50$ . Because the question asks for the proportion below the threshold, this left area is the answer directly, with no subtraction: about 6.68%.
(b) Scale the proportion to the shipment. An expected count is the proportion times the number of units: $500 \times 0.0668 = 33.4$ units. An expected count does not have to be a whole number, so report 33.4 rather than rounding it to a whole laptop.
Answer in context. About 6.68% of units fall under the 10-hour warranty threshold, so in a shipment of 500 the maker should expect about 33.4 units to run short of 10 hours and qualify for replacement.

(a) $z = -1.50$ and the area to the left is 0.0668, about 6.68%; (b) $500 \times 0.0668 = 33.4$ units expected to qualify.

Problem 7

A bakery's daily flour use is approximately normal with mean $\mu = 48$ kg and standard deviation $\sigma = 7$ kg. The owner wants the range that covers the middle 60% of days so she can size a standing order. Between what two amounts do the middle 60% of days fall?

Show the worked solution

State the distribution and split the leftover area. Let $X$ be the flour used on a randomly chosen day, in kilograms, approximately $N(\mu = 48, \sigma = 7)$ . A central 60% leaves $100\% - 60\% = 40\%$ outside, and a normal curve is symmetric, so 20% sits in each tail. The lower boundary therefore has 0.20 to its left and the upper boundary has $0.20 + 0.60 = 0.80$ to its left.
Find both z-scores from the areas. Searching the body of the table for 0.2000 gives 0.2005 at $z = -0.84$ and 0.1977 at $z = -0.85$ , so the nearer entry is $z = -0.84$ . By symmetry the upper boundary is $z = +0.84$ , which the table confirms with 0.7995 to its left, the entry nearest 0.8000.
Unstandardize both boundaries. $(0.84)(7) = 5.88$ kg, so the lower boundary is $x = 48 - 5.88 = 42.12$ kg and the upper boundary is $x = 48 + 5.88 = 53.88$ kg.
Check the width against the empirical rule. The middle 68% of days runs from $48 - 7 = 41$ kg to $48 + 7 = 55$ kg, so a 60% band must sit inside that, and 42.12 to 53.88 does. A band wider than 41 to 55 would mean an arithmetic slip.
Answer in context. On about 60% of days the bakery uses between roughly 42.1 kg and 53.9 kg of flour, so a standing order in that range covers a typical day, and about 20% of days fall short of it while about 20% run over.

$z = \pm 0.84$ , so the middle 60% of days fall between $48 - 5.88 = 42.12$ kg and $48 + 5.88 = 53.88$ kg.

Problem 8

A cereal line fills boxes to a weight that is approximately normal with mean $\mu = 500$ g. Plant records show the 97.5th percentile of fill weight is 514.7 g. (a) Find $\sigma$ . (b) What percent of boxes fall below 490 g? (c) The plant wants to print a minimum weight that only 2% of boxes fall below. What weight should it print? (d) A second line fills smaller boxes with $\mu = 250$ g and $\sigma = 5$ g. A 490 g box from the first line and a 245 g box from the second both look light. Which is more unusual for its own line?

Show the worked solution

(a) State the distribution and translate the percentile. Let $X$ be the fill weight of a randomly chosen box on the first line, in grams, approximately $N(\mu = 500, \sigma)$ with $\sigma$ unknown. The 97.5th percentile has 0.9750 of the area to its left, and the table holds 0.9750 exactly at $z = 1.96$ .
(a) Solve for the standard deviation. Substituting into $x = \mu + z\sigma$ gives $514.7 = 500 + 1.96\sigma$ , so $1.96\sigma = 14.7$ and $\sigma = \frac{14.7}{1.96} = 7.5$ g.
(b) Standardize 490 g with the recovered $\sigma$ . $z = \frac{490 - 500}{7.5} = \frac{-10}{7.5} = -1.3333$ , which rounds to $-1.33$ for a two-decimal table. The area to the left of $z = -1.33$ is 0.0918, about 9.18% of boxes. Technology, keeping $z = -1.3333$ unrounded, returns 0.0912, about 9.12%; the whole gap is the rounding of $z$ , and either value earns credit when the standardized boundary is shown.
(c) Run the calculation backward. A minimum weight that only 2% of boxes fall below has an area of 0.0200 to its left. The table gives 0.0202 at $z = -2.05$ and 0.0197 at $z = -2.06$ , and 0.0202 is nearer to 0.0200, so use $z = -2.05$ . Then $x = 500 + (-2.05)(7.5) = 500 - 15.375 = 484.625$ g, about 484.6 g.
(c) Read the direction of the rounding before printing it. At $z = -2.05$ the printed weight leaves 2.02% of boxes below the label, a hair more than the 2% the plant asked for, so the cautious choice is the next entry down, $z = -2.06$ and $x = 484.55$ g with 1.97% below. Unrounded technology puts the exact cutoff at 484.60 g. All three are within a tenth of a gram, and the exam scores the standardized boundary rather than the last decimal.
(d) Compare across the two lines. Each box has to be standardized inside its own distribution. First line: $z = -1.33$ from part (b). Second line: $z = \frac{245 - 250}{5} = \frac{-5}{5} = -1.00$ . The 490 g box sits 1.33 standard deviations below its line's mean while the 245 g box sits only 1.00 below.
Answer in context. The line's standard deviation is 7.5 g, about 9.18% of boxes fall under 490 g, printing 484.6 g as the minimum leaves only about 2% of boxes short of the label, and the 490 g box is the more unusual of the two because it is farther below its own line's mean in standard deviations, even though it holds far more cereal in grams than the 245 g box.

(a) $\sigma = \frac{14.7}{1.96} = 7.5$ g; (b) $z = -1.33$ , about 9.18% (9.12% with unrounded technology); (c) $z = -2.05$ , so $x = 484.625$ g, about 484.6 g; (d) the 490 g box, at $z = -1.33$ , is more unusual than the 245 g box at $z = -1.00$ .