6.2: Finding Probabilities for the Normal Distribution

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Page ID: 58279

Toros Berberyan, Tracy Nguyen, and Alfie Swan
Citrus College

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Learning Objectives

Convert values to z-scores to compute probabilities using the normal distribution.
Use tables or calculators to find the area under the normal curve representing the likelihood of a value occurring.
Apply the reverse process to determine cut-off scores corresponding to given probabilities.

Finding the Probability of a Normal Distribution

The Empirical Rule is just an approximation and only works for certain values. What if you want to find the probability for x values that are not integer multiples of the standard deviation? The probability is the area under the curve. To find areas under the curve, you need calculus. Before technology, you needed to convert every x value to a standardized number, called the z-score or z-value, or simply just z. The z-score is a measure of how many standard deviations an x value is from the mean. To convert from a normally distributed x value to a z-score, you use the following formula.

Definition $\PageIndex{1}$: z-score

\[z=\dfrac{x-\mu}{\sigma} \label{z-score}\]

where $\mu$= mean of the population of the x value and $\sigma$= standard deviation for the population of the x value

The z-score is normally distributed, with a mean of 0 and a standard deviation of 1. It is known as the standard normal curve. Once you have the z-score, you can look up the z-score in the standard normal distribution table.

Definition $\PageIndex{2}$: standard normal distribution

The standard normal distribution, z, has a mean of $\mu =0$ and a standard deviation of $\sigma =1$.

Standard normal curve. — Figure $\PageIndex{1}$: Standard Normal Curve

Luckily, these days technology can find probabilities for you without converting to the z-score and looking the probabilities up in a table. There are many programs available that will calculate the probability for a normal curve, including Excel and the TI-83/84. There are also online sites available. The following examples show how to do the calculation on the TI-83/84. The command on the TI-83/84 is in the DISTR menu and is normalcdf(. You then type in the lower limit, upper limit, mean, and standard deviation in that order, and including the commas.

Example $\PageIndex{1}$ general normal distribution

The length of a human pregnancy is normally distributed with a mean of 272 days with a standard deviation of 9 days (Bhat & Kushtagi, 2006).

Click on blue links to access the normal distribution tables for negative z-values and positive z-values. After the value is found, press "back" to return to the page.

State the random variable.
Find the probability that a pregnancy lasts more than 280 days.
Find the probability that a pregnancy lasts less than 250 days.
Find the probability that a pregnancy lasts between 265 and 280 days.
Find the cutoff length of pregnancy such that the shortest 10% of all pregnancies fall below this value.
Suppose you meet a woman who says that she was pregnant for less than 250 days. Would this be unusual, and what might you think?

Solution

a. x = length of a human pregnancy

b. First, translate the statement into a mathematical statement.

P (x>280)

Now, draw a picture. Remember that the center of this normal curve is 272.

Normal distribution with shaded area to the right of 280. — Figure $\PageIndex{2}$: Normal Distribution With Shaded Area to the Right of 280.

To find the probability on the TI-83/84, looking at the picture, you realize the lower limit is 280. The upper limit is infinity. The calculator doesn’t have infinity on it, so you need to put in a really big number. Some people like to put in 1000, but if you are working with numbers that are bigger than 1000, then you would have to remember to change the upper limit. The safest number to use is $1 \times 10^{99}$, which you put in the calculator as 1E99 (where E is the EE button on the calculator). The command looks like this:

$\text{normalcdf}(280,1 E 99,272,9)$

TI-83/84 output of area to the right of 280. — Figure $\PageIndex{3}$: TI-83/84 Output of Area to the Right of 280.

Thus, $P(x>280) \approx 0.187$

Thus 18.7% of all pregnancies last more than 280 days. This is not unusual since the probability is greater than 5%.

To find the probability using a z-table, we need to first compute the z-score and then use the table to find the probability. Based on the problem, we know that $\mu$ (population mean) is 272 days and $\sigma$ (population standard deviation) is 9 days. We know that we are finding the probability of a pregnancy lasting more than 280 days, so x our random variable x will be 280 days.

$z=\dfrac{x-\mu}{\sigma}$

$z=\dfrac{280-272}{9} $

$z=\dfrac{8}{9}$

$z=0.89$ (rounded to two decimal places)

As a result, P (x >280) = P(z > 0.89). Since z is more than 0.89, our shaded area will be on the right of z = 0.89.

We can use the z-table to find our probability to the left. If we convert the normal distribution to the standard normal distribution, we get the following:

Normal distribution where z = 0.89 and the the shaded area is on the right. — Figure $\PageIndex{4}$: Normal Distribution, Z = 0.89, Shaded on the Right

From the z-table, it only gives us the probability to the left of z-score. Since we know the total area of the normal distribution is 1. To get the area to the right, we need to take 1 minus the area to the left. The work is provided below.

P(z > 0.89) = 1 - P(z < 0.89)

To find the area or probability to the left of z = 0.89, we use the standard normal distribution table provided below. We go down 0.8 and move over to the right 0.09 to get the area of 0.8133. This area (probability) represents the area to the left of z=0.89.

Table $\PageIndex{1}$: Normal Distribution Table For Positive Z-values

Based on the z-table above, the Left Area will be 0.8133.

Right Area = 1 - Left Area

Right Area = 1 - 0.8133 = 0.1867

P(z>0.89)= 0.1867 = 18.67%

Using the table, P(x >280) = 18.67%. This means that the probability that a pregnancy will last more than 280 days is 18.67%.

The answer using the z-table is slightly off due to rounding our z-value to two decimal places. The calculator uses more decimal places for z, hence it has a more accurate answer.

c. First, translate the statement into a mathematical statement.

P (x<250)

Now, draw a picture. Remember that the center of this normal curve is 272.

Graph of normal distribution with area to the left of 250. — Figure $\PageIndex{5}$: Graph of Normal Distribution with Area to the Left of 250.

To find the probability on the TI-83/84, looking at the picture, though it is hard to see in this case, the lower limit is negative infinity. Again, the calculator doesn’t have this on it, put in a really small number, such as $-1 \times 10^{99}=-1 E 99$ on the calculator.

TI-83/84 output for area to the left of 250. — Figure $\PageIndex{6}$: TI-83/84 Output for Area to the Left of 250.

$P(x<250)=\text { normalcdf }(-1 E 99,250,272,9)=0.0073$

Thus, 0.73% of all pregnancies last less than 250 days. This is unusual since the probability is less than 5%.

To find the probability using a z-table, we need to first compute the z-score and then use the table to find the probability. Based on the problem, we know that $\mu$ (population mean) is 272 days and $\sigma$ (population standard deviation) is 9 days. We know that we are finding the probability of a pregnancy lasting less than 250 days, so x our random variable x will be 250 days.

$z=\dfrac{x-\mu}{\sigma}$

$z=\dfrac{250-272}{9} $

$z=\dfrac{-22}{9}$

$z=-2.44$ (rounded to two decimal places)

As a result, P (x < 250) = P(z < -2.44). Since z is less than -2.44, our shaded area will be on the left of z = -2.44.

We can use the z-table to find our probability to the left. If we convert the normal distribution to the standard normal distribution, we get the following:

Normal distribution where z = -2.44 and the the shaded area is on the left. — Figure $\PageIndex{7}$: Normal Distribution, Z = -2.44, Shaded on the Left

We use the standard normal distribution table provided below to find the area or probability to the left of z = -2.44. We go down -2.4 and move over to the right 0.04 to get the area of 0.0073. This area (probability) represents the area to the left of z=-2.44.

Table $\PageIndex{2}$: Normal Distribution Table For Negative Z-values

Based on the z-table above, the Left Area will be 0.0073, which is the probability that z is less than -2.44. If we express the answer as a percentage, we get the following answer:

P( z < -2.44) = 0.73%

Using the table, P(x <250) = 0.73%. This means the probability that a pregnancy will last less than 250 days is 0.73%.

The answer using the z-table is slightly off due to rounding our z-value to two decimal places. The calculator uses more decimal places for z, hence it has a more accurate answer.

d. First translate the statement into a mathematical statement.

P (265<x<280)

Now, draw a picture. Remember that the center of this normal curve is 272.

Normal distribution with area between 265 and 280. — Figure $\PageIndex{8}$: Normal Distribution with Area Between 265 and 280.

In this case, the lower limit is 265 and the upper limit is 280.

Using the calculator

TI-83/84 output for area between 265 and 280. — Figure $\PageIndex{9}$: TI-83/84 Output for Area Between 265 and 280.

$P(265<x<280)=\text { normalcdf }(265,280,272,9)=0.595$

Thus 59.5% of all pregnancies last between 265 and 280 days.

To find the probability using a z-table, we need to first compute the z-scores and then use the table to find the probability. Based on the problem, we know that $\mu$ (population mean) is 272 days and $\sigma$ (population standard deviation) is 9 days. We know that we are finding the probability of a pregnancy lasting between 265 and 280 days, so x our random variable will be 265 days and 280 days. For each x-value, we need to find their corresponding z-value.

$z=\dfrac{x-\mu}{\sigma}$

$z=\dfrac{265-272}{9} $

$z=\dfrac{-7}{9}$

$z=-0.78$ (rounded to two decimal places)

$z=\dfrac{x-\mu}{\sigma}$

$z=\dfrac{280-272}{9} $

$z=\dfrac{8}{9}$

$z=0.89$ (rounded to two decimal places)

As a result, P(265 < x < 280 ) = P(-0.78 < z < 0.89).

Normal distribution where shaded area is between z = -0.78 and z = 0.89. — Figure $\PageIndex{10}$: Normal Distribution Where Shaded Area is Between z = -0.78 and z = 0.89

We can find our probability to the left of each z-score by using the z-table. If we subtract the higher probability value with the lower probability value, we will find the area (probability) between z = -0.78 and z = 0.89.

$P\left(-0.78<z<0.89\right)=P\left(z<0.89\right)-P\left(z<-0.78\right)$

To find $P\left(z<0.89\right)$ and $P\left(z<-0.78\right)$ we use the z-table.

z = 0.89

$P\left(z<0.89\right)=0.8133$

Table $\PageIndex{1}$: Normal Distribution Table For Positive Z-values

z = -0.78

To find the area or probability to the left of z = -0.78, we use the standard normal distribution table provided below. We go down -0.7 and move over to the right 0.08 to get the area of 0.2177. This area (probability) represents the area to the left of z=-0.78.

$P\left(z<-0.78\right)=0.2177$

Table $\PageIndex{2}$: Normal Distribution Table For Negative Z-values

Now that we found $P\left(z<0.89\right)=0.8133$ and $P\left(z<-0.78\right)=0.2177$, we can use the following formula:

$P\left(265<x<280\right)=P\left(-0.78<z<0.89\right)=P\left(z<0.89\right)-P\left(z<-0.78\right)$

$P\left(265<x<280\right)=P\left(-0.78<z<0.89\right)=0.8133-0.2177$

$P\left(265<x<280\right)=P\left(-0.78<z<0.89\right)=0.5956$

Thus, the probability of a pregnancy lasting between 265 and 280 days will be 59.56%.

e. This problem asks you to find an x value from a probability. You want to find the x value that has 10% of the length of pregnancies to the left of it. On the TI-83/84, the command is in the DISTR menu and is called invNorm(. The invNorm( command needs the area to the left. In this case, that is the area you are given. For the command on the calculator, once you have invNorm( on the main screen you type in the probability to the left, mean, standard deviation, in that order with the commas.

TI-83/84 output for inverse norm function — Figure $\PageIndex{11}$: TI-83/84 Output for Inverse Norm Function

Thus, 10% of all pregnancies last less than approximately 260 days.

f. From part (c), you found the probability that a pregnancy lasts less than 250 days is 0.73%. Since this is less than 5%, it is very unusual. You would think that either the woman had a premature baby, or that she may be wrong about when she became pregnant.

Example $\PageIndex{2}$ general normal distribution

The mean mathematics SAT score in 2012 was 514 with a standard deviation of 117 ("Total group profile," 2012). Assume the mathematics SAT score is normally distributed.

Click on blue links to access the normal distribution tables for negative z-values and positive z-values. After the value is found, press "back" to return to the page.

State the random variable.
Find the probability that a person has a mathematics SAT score over 700.
Find the probability that a person has a mathematics SAT score of less than 400.
Find the probability that a person has a mathematics SAT score between 500 and 650.
Find the mathematics SAT score that represents the top 1% of all scores.

Solution

a. x = mathematics SAT score

b. First, translate the statement into a mathematical statement.

P (x>700)

Now, draw a picture. Remember that the center of this normal curve is 514.

Normal distribution with area to the right of 700. — Figure $\PageIndex{12}$: Normal Distribution with Area to the Right of 700

On TI-83/84: $P(x>700)=\text { normalcdf }(700,1 E 99,514,117) \approx 0.056$

There is a 5.6% chance that a person scored above 700 on the mathematics SAT. This is not unusual.

c. First, translate the statement into a mathematical statement.

P (x<400)

Now, draw a picture. Remember that the center of this normal curve is 514.

Normal distribution with area to the left of 400. — Figure $\PageIndex{13}$: Normal Distribution with Area to the Left of 400.

On TI-83/84: $P(x<400)=\text { normalcdf }(-1 E 99,400,514,117) \approx 0.165$

So, there is a 16.5% chance that a person scores less than 400 on the mathematics part of the SAT.

d. First translate the statement into a mathematical statement.

P (500<x<650)

Now, draw a picture. Remember that the center of this normal curve is 514.

Normal distribution of area between 500 and 650. — Figure $\PageIndex{14}$: Normal Distribution of Area Between 500 and 650.

On TI-83/84: $P(500<x<650)=\text { normalcdf }(500,650,514,117) \approx 0.425$

So, there is a 42.5% chance that a person has a mathematical SAT score between 500 and 650.

e. This problem asks you to find an x value from a probability. You want to find the x value that has 1% of the mathematics SAT scores to the right of it. Remember, the calculator and R always need the area to the left; you need to find the area to the left by 1 - 0.01 = 0.99.

On TI-83/84: $\text{invNorm}(.99,514,117) \approx 786$

So, 1% of all people who took the SAT scored over 786 points on the mathematics SAT.

Finding the Minimum and Maximum of a Random Variable for a Normal Distribution

When working with a normal distribution, if we want to know the minimum value to be part of the top range, the maximum value to be in the bottom range, or the minimum or maximum value for the middle range, then we can use the invNorm() function to obtain the answers.

Example $\PageIndex{1}$ Find the Minimum and Maximum of the Random Variable

The ages of students at Citrus College are normally distributed with a mean of 24 years and a standard deviation of 3 years.

Click on blue links to access the normal distribution tables for negative z-values and positive z-values. After the value is found, press "back" to return to the page.

What is the maximum age a student can be to fall in the bottom 20% of the age distribution at Citrus College?
What is the minimum age a student must be to fall in the top 20% of the age distribution at Citrus College?
What are the minimum and maximum ages that encompass the middle 50% of the age distribution at Citrus College?

Solution

a. To find the maximum age to be part of the bottom 20%, we draw a bell-shaped curve and shade the bottom 20% of the graph.

Normal distribution with bottom 20% shaded where the fence is the maximum value. — Figure $\PageIndex{15}$: Normal distribution with the bottom 20% shaded, where the fence is the maximum value.

Calculating z Using invNorm() function:

To find the z-value with a graphing calculator, use the invNorm() function.

Press the 2nd button.
Press the VARS button to open the "DISTR" menu.
Select 2: invNorm().
Use the invNorm(area to the left of z, mean, standard deviation) function and press the ENTER button. If done correctly, the screen will show the following:

Inverse norm function from TI-84+ output of 21.475. — Figure $\PageIndex{16}$: On graphing calculator screen, displays invNorm(0.20, 24, 3) and its calculation of 21.4751363

As a result, the maximum age to be part of the bottom 20% of Citrus College students will be roughly 21.5 years old.

b. To find the maximum age to be part of the top 20%, we draw a bell-shaped curve and shade the top 20% of the graph. This will be on the right side.

Figure $\PageIndex{17}$: Normal distribution with top 20% shaded where the fence is the minimum value.

Calculating z Using invNorm() function:

To find the z-value with a graphing calculator, use the invNorm() function.

Press the 2ND button.
Press the VARS button to open the "DISTR" menu.
Select 2: invNorm().
Use the invNorm(area to the left of z, mean, standard deviation) function and press the ENTER button. In this case, the right is 20% so the area to the left of z will be 80%. If done correctly, the screen will show the following:

Figure $\PageIndex{18}$: On graphing calculator screen, displays invNorm(0.80, 24, 3) and its calculation of 26.5248637

As a result, the minimum age to be part of the top 20% of Citrus College students will be roughly 26.5 years old.

c. To find the minimum and maximum ages that encompass the middle 50% of the age distribution at Citrus College, we draw a bell-shaped curve and shade the middle 50% of the graph.

Normal distribution with middle 50%. — Figure $\PageIndex{19}$: Normal distribution with middle 50% shaded, where we have a minimum and maximum vertical fences on the sides.

Calculating z Using invNorm() function:

To find the z-value with a graphing calculator, use the invNorm() function.

Press the 2ND button.
Press the VARS button to open the "DISTR" menu.
Select 2: invNorm().
Use the invNorm(area to the left of z, mean, standard deviation) function to find the minimum value. In order to find the minimum value, the area to the left of the fence is what is needed. The whole graph should add up to 100%. The area to the left and right is symmetric. As a result, the outside areas have to be 25%. Here is an image to illustrate this idea.

Figure $\PageIndex{20}$: Normal distribution graph where the middle region is 50% and the side regions are 25%.

As a result, the area to the left of the minimum fence ($x_1$) will be 25%. The area to the left of the maximum fence ($x_2$) will be the middle region (50%) plus the left region (25%) which will be 75%. These are the values we put for the area in decimal form inside invNorm() function. The mean for both will still be 24 and standard deviation will be 3. When done inputting the invNorm() function on the calculator, press the ENTER button. If done correctly, the screen will show the following:

Inverse norm function on TI-84+ calculator with outputs 21.977 and 26.023. — Figure $\PageIndex{21}$: On graphing calculator screen, displays shows invNorm(0.25, 24, 3) and its calculation of 21.97653075. It also shows invNorm(0.75, 24, 3) and its calculation of 26.02346925.

As a result, the middle 50% of the age distribution at Citrus College is roughly between 22 and 26 years old.

Authors

"6.2: Finding Probabilities for the Normal Distribution" by Toros Berberyan, Tracy Nguyen, and Alfie Swan is licensed under CC BY-SA 4.0

Attributions

"6.3: Finding Probabilities for the Normal Distribution" by Kathryn Kozak is licensed under CC BY-SA 4.0

Exercises

The size of fish is very important to commercial fishing. A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm (Ovegard, Berndt & Lunneryd, 2012). Assume the length of the fish is normally distributed. Find the probability that an Atlantic cod has a length of less than 52 cm. Use the standard normal distribution chart, round the z-value to two decimals, and round the answer to four decimals.

Scan the QR code or click on it to open the MyOpenMath version of the above question with step-by-step guidance.
MyOpenMath is a free online learning platform designed to support math instruction through automated homework, quizzes, and assessments. You must register for MyOpenMath and sign in to view the question.

The size of fish is very important to commercial fishing. A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm (Ovegard, Berndt & Lunneryd, 2012). Assume the length of the fish is normally distributed. Find the probability that an Atlantic cod has a length of more than 54 cm. Use the standard normal distribution chart, round the z-value to two decimals, and round the answer to four decimals.

The size of fish is very important to commercial fishing. A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm (Ovegard, Berndt & Lunneryd, 2012). Assume the length of the fish is normally distributed. Find the probability that an Atlantic cod has a length between 47.5 and 55.5 cm. Use the standard normal distribution chart, round the z-value to two decimals, and round the answers to four decimals.

In the United States, males between the ages of 40 and 49 eat on average 103.1 g of fat every day with a standard deviation of 4.32 g ("What we eat," 2012). Assume that the amount of fat a person eats is normally distributed. Find the probability that a man aged 40-49 in the U.S. eats between 90 and 105 g of fat daily. Use the standard normal distribution chart, round the z-value to two decimals, and round the answers to four decimals.

In the United States, males between the ages of 40 and 49 eat on average 103.1 g of fat every day with a standard deviation of 4.32 g ("What we eat," 2012). Assume that the amount of fat a person eats is normally distributed. Find the probability that a man aged 40-49 in the U.S. eats greater than 110 g of fat daily. Use the standard normal distribution chart, round the z-value to two decimals, and round the answer to four decimals.

In the United States, males between the ages of 40 and 49 eat on average 103.1 g of fat every day with a standard deviation of 4.32 g ("What we eat," 2012). Assume that the amount of fat a person eats is normally distributed. Find the probability that a man aged 40-49 in the U.S. eats less than 93 g of fat daily. Use the standard normal distribution chart, round the z-value to two decimals, and round the answer to four decimals.

The mean starting salary for nurses is $67,694 nationally ("Staff nurse -," 2013). The standard deviation is approximately $10,333. Assume that the starting salary is normally distributed. Find the probability that a starting nurse will make more than $80,000. Round the answer to four decimals.

The mean starting salary for nurses is $67,694 nationally ("Staff nurse -," 2013). The standard deviation is approximately $10,333. Assume that the starting salary is normally distributed. Find the probability that a starting nurse will make between $55,000 and $72,000. Round the answers to four decimals.

The mean starting salary for nurses is $67,694 nationally ("Staff nurse -," 2013). The standard deviation is approximately $10,333. Assume that the starting salary is normally distributed. Find the probability that a starting nurse will make less than $60,000. Round the answer to four decimals.

In a large statistics class, the scores on Exam 2 are normally distributed with a mean of 74 and a standard deviation of 8. What is the minimum score a student must earn to be in the top 10% of the class? Use the standard normal distribution chart, round the z-value to two decimals, and round the answer to a whole number.

At a college cafeteria, the price of lunch is normally distributed with a mean of $9.50 and a standard deviation of $1.75. What is the range of lunch prices that includes the middle 40% of all lunch prices? Use the standard normal distribution chart, round the z-value to two decimals, and round the answers to two decimal places.

Answers

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Definition \(\PageIndex{1}\): z-score

Definition \(\PageIndex{2}\): standard normal distribution

Example \(\PageIndex{1}\) general normal distribution

Solution

Right Area = 1 - Left Area

z = 0.89

z = -0.78

Example \(\PageIndex{2}\) general normal distribution

Solution

Example \(\PageIndex{1}\) Find the Minimum and Maximum of the Random Variable

Solution

Calculating z Using invNorm() function:

Calculating z Using invNorm() function:

Calculating z Using invNorm() function: