Example 1

What is the z-score of a value of 27, given a set mean of 24, and a standard deviation of 2?

Solution

To find the z-score we need to divide the difference between the value, 27, and the mean, 24, by the standard deviation of the set, 2. [latex-display]\displaystyle\text{z-score}=\frac{27-\mu}{\sigma}=\frac{27-24}{2}=\frac{3}{2}\\[/latex-display] [latex-display]\displaystyle\text{z-score of }27=+1.5\\[/latex-display] This indicates that 27 is 1.5 standard deviations above the mean.

Example 2

What is the z-score of a value of 104.5, in a set with µ = 125 and σ = 6.2?

Solution

Find the difference between the given value and the mean, then divide it by the standard deviation. [latex-display]\displaystyle\text{z-score}=\frac{104.5-\mu}{\sigma}=\frac{104.5-125}{6.2}=\frac{-20.5}{6.2}\\[/latex-display] [latex-display]\displaystyle\text{z-score of }104005=-3.306\\[/latex-display] Note that the z-score is negative, since the measured value, 104.5, is less than (below) the mean, 125.

Example 3

Find the value represented by a z-score of 2.403, given µ = 63 and σ = 4.25.

Solution

This one requires that we solve for a missing value rather than for a missing z-score, so we just need to fill in our formula with what we know and solve for the missing value: [latex-display]\displaystyle\text{z-score}=\frac{x-\mu}{\sigma}\\[/latex-display] [latex-display]\displaystyle2.403=\frac{x-63}{4.25}\\[/latex-display] [latex-display]\displaystyle10.213=x-63\\[/latex-display] [latex-display]73.213=x\\[/latex-display] 73.213 has a z-score of 2.403

Guided Practice

1. What is the z-score of the price of a pair of skis that cost $247, if the mean ski price is $279, with a standard deviation of $16?
2. What is the z-score of a 5-scoop ice cream cone if the mean number of scoops is 3, with a standard deviation of 1 scoop?
3. What is the z-score of the weight of a cow that tips the scales at 825 lbs, if the mean weight for cows of her type is 1150 lbs, with a standard deviation of 77 lbs?
4. What is the z-score of a measured value of 0.0034, given µ = 0.0041 and σ = 0.0008?

Solutions

1. First find the difference between the measured value and the mean, then divide that difference by the standard deviation: [latex-display]\frac{247-279}{16}=\frac{-32}{16}\\[/latex-display] z-score = −2
2. This one is easy: The difference between 5 scoops and 3 scoops is +2, and we divide that by the standard deviation of 1, so the z-score is +2.
3. First find the difference between the measured value and the mean, then divide that difference by the standard deviation: [latex-display]\frac{825\text{ lbs}-1150\text{ lbs}}{771\text{ lbs}}=\frac{-325}{77}\\[/latex-display] z-score = −4.2407
4. First find the difference between the measured value and the mean, then divide that difference by the standard deviation: [latex-display]\frac{0.0034-0.0041}{0.0008}=\frac{-0.0007}{0.0008}\\[/latex-display] z-score = −0.875

Example 4

What is the probability that a value with a z-score less than 2.47 will occur in a normal distribution?

Solution

Scroll up to the table above and find “2.4” on the left or right side. Now move across the table to “0.07” on the top or bottom, and record the value in the cell: 0.9932. That tells us that 99.32% of values in the set are at or below a z-score of 2.47.

Example 5

What is the probability that a value with a z-score greater than 1.53 will occur in a normal distribution?

Solution

Scroll up to the table of z-score probabilities again and find the intersection between 1.5 on the left or right and 3 on the top or bottom, record the value in the cell: 0.937. That decimal lets us know that 93.7% of values in the set are below the z-score of 1.53. To find the percentage that is above that value, we subtract 0.937 from 1.0 (or 93.7% from 100%), to get 0.063 or 6.3%.

Example 6

What is the probability of a random selection being less than 3.65, given a normal distribution with µ = 5 and σ = 2.2?

Solution

This question requires us to first find the z-score for the value 3.65, then calculate the percentage of values below that z-score from a reference.

1. Find the z-score for 3.65, using the z-score formula: [latex]\frac{(x-\mu)}{\sigma}=\frac{3.65-5}{2.2}=\frac{-1.35}{2.2}\approx{-0.61}\\[/latex]
2. Now we can scroll up to our z-score reference above and find the intersection of 0.6 and 0.01, which should be .7291.
3. Since this is a negative z-score, and we want the percentage of values below it, we subtract that decimal from 1.0 (reference the three steps highlighted by bullet points below the chart if you didn’t recall this), to get 1 − .7291 = .2709

There is approximately a 27.09% probability that a value less than 3.65 would occur from a random selection of a normal distribution with mean 5 and standard deviation 2. 2.

Guided Practice

1. What is the probability of occurrence of a value with z-score greater than 1.24?
2. What is the probability of z < −.23?
3. What is P(Z < 2.13)?

Solutions

1. Since this is a positive z-score, we can use the value for z = 1.24 directly from the table, and just express it as a percentage: 0.8925 or 89.25%
2. This is a negative z-score, and we want the percentage of values greater than it, so we need to subtract the value for z = +0.23 from 1: 1 − 0.591 = .409 or 40.9%
3. This is a positive z-score, and we need the percentage of values below it, so we can use the percentage associated with z = +2.13 directly from the table: 0.9834 or 98.34%

Example 7

What is the probability associated with a z-score between 1.2 and 2.31?

Solution

To evaluate the probability of a value occurring within a given range, you need to find the probability of both the upper and lower values in the range, and subtract to find the difference.

• First find z = 1.2 on the z-score probability reference above: .8849. Remember that value represents the percentage of values below 1.2.
• Next, find and record the value associated with z = 2.31: .9896
• Since approximately 88.49% of all values are below z = 1.2 and approximately 98.96% of all values are below z = 2.31, there are 98.96% −88.49% = 10.47% of values between.

Example 8

What is the probability that a value with a z-score between -1.32 and +1.49 will occur in a normal distribution?

Solution

Let’s use the online calculator at mathportal.org for this one. When you open the page, you should see a window like this: All you need to do is select the radio button to the left of the first type of probability, input “−1.32” into the first box, and 1.49 into the second. When you click Compute, you should get the result P(−1.32 < Z < 1.49) = 0.8385 Which tells us that there is approximately and 83.85% probability that a value with a z-score between 1.32 and 1.49 will occur in a normal distribution. Notice that the calculator also details the steps involved with finding the answer:

1. Estimate the probability using a graph, so you have an idea of what your answer should be.
2. Find the probability of z < 1.49, using a reference. (0.9319)
3. Find the probability of z < −1.32, again, using a reference. (0.0934)
4. Subtract the values: 0.9319 − 0.0934 = 0.8385 or 83.85%

Example 9

What is the probability that a random selection will be between 8.45 and 10.25, if it is from a normal distribution with µ = 10 and σ = 2?

Solution

This question requires us to first find the z-scores for the value 8.45 and 10.25, then calculate the percentage of value between them by using values from a z-score reference and finding the difference.

1. Find the z-score for 8.45, using the z-score formula: [latex]\frac{(x-\mu)}{\sigma}\\[/latex] [latex]\frac{8.45-10}{2}=\frac{-1.55}{2}\approx{-0.78}\\[/latex]
2. Find the z-score for 10.25 the same way: [latex]\frac{10.25-10}{2}=\frac{0.25}{2}\approx{.13}\\[/latex]
3. Now find the percentages for each, using a reference (don’t forget we want the probability of values less than our negative score and less than our positive score, so we can find the values between): P(Z < −0.78) = .2177 or 21.77% P(Z < .13) = .5517 or 55.17%
4. At this point, let’s sketch the graph to get an idea what we are looking for:
5. Finally, subtract the values to find the difference: .5517 − .2177 = .3340 or about 33.4% There is approximately a 33.4% probability that a value between 8.45 and 10.25 would result from a random selection of a normal distribution with mean 10 and standard deviation 2.

Guided Practice

1. What is the probability of a z-score between −0.93 and 2.11?
2. What is P(1.39 < Z < 2.03)?
3. What is P(−2.11 < Z < 2.11)?

Solutions

1. Using the z-score probability table above, we can see that the probability of a value below −0.93 is .1762, and the probability of a value below 2.11 is .9826. Therefore, the probability of a value between them is .9826−.1762 = .8064 or 80.64%
2. Using the z-score probability table, we see that the probability of a value below z = 1.39 is .9177, and a value below z = 2.03 is .9788. That means that the probability of a value between them is .9788 − .9177 = .0611 or 6.11%
3. Using the online calculator at mathportal.org, we select the top calculation with the associated radio button to the left of it, enter “−2.11” in the first box, and “2.11” in the second box. Click Compute to get “ .9652,” and convert to a percentage. The probability of a z-score between −2.11 and +2.11 is about 96.52%.