Lesson Explainer: Applications of Normal Distribution Mathematics

In this explainer, we will learn how to apply the normal distribution in real-life situations.

Many continuous variables in the real world approximately follow the normal distribution. Variables like heights and weights collected from unbiased samples are expected to be normally distributed. If we collect the values of such variables from a large random sample, then we expect the distribution to resemble the following histogram.

If the distribution of a continuous variable is symmetric and concentrated near the mean (like the data set pictured above), then we can assume that the variable is approximately normally distributed. To approximate the percentage of data points lying within a given range in such variables, we can use the normal probability distribution.

For example, say that we want to approximate the percentage of people from France whose heights are between 160 cm and 180 cm. Assume that, after collecting the data from a large random sample, we have computed the mean height of the sample to be 175 cm and the standard deviation to be 5 cm. We would begin this problem by defining a normal random variable 𝑋 with mean 175 cm and standard deviation 5 cm. We recall that π‘‹βˆΌπ‘ο€Ήπœ‡,πœŽο…οŠ¨ denotes that the variable 𝑋 is normally distributed with mean πœ‡ and standard deviation 𝜎. Using this notation, π‘‹βˆΌπ‘ο€Ή175,5ο…οŠ¨. Then, the percentage of people from France whose heights are between 160 cm and 180 cm is approximated by the probability 𝑃(160≀𝑋≀180). We remember that since 𝑋 is a continuous random variable, the strict inequality < and the weak inequality ≀ are interchangeable. This is because the probability that 𝑋 will take a particular value is zero; that is, 𝑃(𝑋=π‘₯)=0 for any π‘₯. In particular, this probability can also be written in several different ways: 𝑃(160≀𝑋≀180)=𝑃(160<𝑋≀180)=𝑃(160≀𝑋<180)=𝑃(160<𝑋<180).

So, we do not need to be concerned about whether or not the phrase β€œbetween 160 cm and 180 cm” includes the endpoints 160 cm and 180 cm.

Recall that if π‘‹βˆΌπ‘ο€Ήπœ‡,πœŽο…οŠ¨, then 𝑍=π‘‹βˆ’πœ‡πœŽ is the standard normal variable π‘βˆΌπ‘ο€Ή0,1ο…οŠ¨. Then, the probability for 𝑍 is obtained using the bell curve and the standard normal table. In this explainer, we will use the standard normal table that provides probabilities of the form 𝑃(0≀𝑍≀𝑧) for 𝑧β‰₯0.

To compute the probability 𝑃(160≀𝑋≀180), we begin by standardizing the normal distribution: 𝑃(160≀𝑋≀180)=𝑃(160βˆ’175β‰€π‘‹βˆ’πœ‡β‰€180βˆ’175)=𝑃(βˆ’15β‰€π‘‹βˆ’πœ‡β‰€5)=π‘ƒο€½βˆ’155β‰€π‘‹βˆ’πœ‡πœŽβ‰€55=π‘ƒο€½βˆ’3β‰€π‘‹βˆ’πœ‡πœŽβ‰€1.

Since 𝑍=π‘‹βˆ’πœ‡πœŽ is the standard normal random variable, we analyze the region {βˆ’3≀𝑍≀1} by drawing the bell curves.

Then, we can write 𝑃(βˆ’3≀𝑍≀1)=𝑃(0≀𝑍≀3)+𝑃(0≀𝑍≀1).

From the standard normal table, we find 𝑃(0≀𝑍≀3)=0.4987 and 𝑃(0≀𝑍≀1)=0.3413. Summing these numbers, we get 𝑃(βˆ’3≀𝑍≀1)=0.4987+0.3413=0.84.

So 𝑃(160≀𝑋≀180)=0.84. We remember that, to convert a probability to a percentage, we need to multiply the probability by 100. So, 0.84Γ—100=84%, which means that the heights of approximately 84% of people from France are between 160 cm and 180 cm.

Given a value π‘₯ of a random variable, the 𝑧-score represents its position relative to the mean value, measured by the number of standard deviations.

Definition: 𝑧-Score

Let π‘₯ be a data point from a variable with mean πœ‡ and standard deviation 𝜎. Then, the 𝑧-score associated with π‘₯ is given by 𝑧=π‘₯βˆ’πœ‡πœŽ.

The probability associated with a 𝑧-score is 𝑃(𝑍≀𝑧), where 𝑍 is the standard normal variable.

We note that the formula above is analogous to that of standardizing a normal distribution, except that both 𝑧 and π‘₯ are in lowercase. For example, a 𝑧-score of βˆ’1.56 indicates that the value π‘₯ is 1.56Γ—πœŽ to the left of πœ‡. In other words, π‘₯=πœ‡βˆ’1.56Γ—πœŽ.

Let us look at a few examples to familiarize ourselves with different contexts.

Example 1: Estimating Normal Distribution Probabilities in Context

A crop of apples has a mean weight of 105 g and a standard deviation of 3 g. It is assumed that a normal distribution is an appropriate model for this data. What is the approximate probability that a randomly selected apple from the crop has a weight less than 105 g?

Answer

We begin by using 𝑋 to represent the weight of an apple, which is assumed to follow the normal distribution with mean πœ‡=105 and standard deviation 𝜎=3. In other words, we can write π‘‹βˆΌπ‘ο€Ή105,3ο…οŠ¨. We need to compute the probability 𝑃(𝑋<105).

To standardize the normal distribution, we first subtract πœ‡=105 from each side. Then, we divide each side by 𝜎=3. Lastly, we replace 𝑍 with π‘‹βˆ’πœ‡πœŽ: 𝑃(𝑋<105)=𝑃(π‘‹βˆ’πœ‡<0)=π‘ƒο€½π‘‹βˆ’πœ‡πœŽ<0=𝑃(𝑍<0).

By symmetry of the bell curve, 𝑃(𝑍<0)=0.5, as seen in the picture below.

We remember that, to convert a probability to a percentage, we need to multiply the probability by 100. So 0.5Γ—100=50%, which means that the probability that a randomly selected apple from the crop has a weight less than 105 g is 50%.

In the last example, we presented the process of standardizing the normal distribution to compute its probability. However, this was unnecessary for this particular example, since we are asked simply to compute the probability that a randomly selected apple had a weight less than the mean. Since the weights of apples are assumed to be normally distributed, this means in particular that the distribution is symmetric about the mean. In other words, approximately half of the apples would have weights less than the mean, and the other half will have weights above it. Using this reasoning, we could have inferred straight away that 𝑃(𝑋<πœ‡)=0.5.

In our next example, we will demonstrate the process for computing the probability for a nontrivial region.

Example 2: Calculating Probabilities from a Normal Distribution in Context

The monthly salaries of workers at a factory are normally distributed with mean 210 pounds and standard deviation 10 pounds. Determine the probability of choosing at random a worker with a salary between 184 and 233 pounds.

Answer

Let 𝑋 represent the monthly salary, which is normally distributed with πœ‡=210 and 𝜎=10. We need to compute 𝑃(184≀𝑋≀233). Standardizing the normal distribution, 𝑃(184≀𝑋≀233)=𝑃(βˆ’26β‰€π‘‹βˆ’πœ‡β‰€23)=π‘ƒο€½βˆ’2610β‰€π‘‹βˆ’πœ‡πœŽβ‰€2310=𝑃(βˆ’2.6≀𝑍≀2.3).

Since 𝑃(βˆ’2.6≀𝑍≀2.3) involves positive and negative values of 𝑍, we need to split this into the positive and the negative regions. Let us think through the process using pictures of the bell curve.

This leads to the following equations: 𝑃(βˆ’2.6≀𝑍≀2.3)=𝑃(βˆ’2.6≀𝑍≀0)+𝑃(0≀𝑍≀2.3)=𝑃(0≀𝑍≀2.6)+𝑃(0≀𝑍≀2.3).

Using the standard normal table, we get 𝑃(0≀𝑍≀2.6)=0.4953 and 𝑃(0≀𝑍≀2.3)=0.4893. Adding up the probabilities, 𝑃(βˆ’2.6≀𝑍≀2.3)=𝑃(0≀𝑍≀2.6)+𝑃(0≀𝑍≀2.3)=0.4953+0.4893=0.9846.

So, the probability of choosing at random a worker with a salary between 184 and 233 pounds is 0.9846.

In the next two examples, we will consider the percentage of data lying within a given range. We remember that the probability is converted into the percentage after multiplying by 100.

Example 3: Estimating Population Percentages from a Normal Distribution in Context

The masses of a population of blackbirds are normally distributed with mean 103 g and standard deviation 11 g.

  1. To the nearest integer, what percentage of blackbirds have masses less than 110 g?
  2. To the nearest tenth, what percentage of blackbirds have masses greater than 124 g?
  3. To the nearest integer, what percentage of blackbirds have masses between 95 g and 120 g?

Answer

Let 𝑋 be the mass of a blackbird. Then, π‘‹βˆΌπ‘ο€Ή103,11ο…οŠ¨.

Part 1

Let us find the percentage of blackbirds with masses less than 110 g. Using probability notation, we need to compute 𝑃(𝑋<110). We being by standardizing the normal distribution: 𝑃(𝑋<110)=𝑃(π‘‹βˆ’πœ‡<7)=π‘ƒο€½π‘‹βˆ’πœ‡πœŽ<711=𝑃𝑍<711.

To use the standard normal table, we need to round 711 to the nearest hundredth, 0.64. Then, we split 𝑃(𝑍<0.64) into the positive and the negative regions as pictured below.

Then, 𝑃(𝑍<0.64)=𝑃(𝑍≀0)+𝑃(0≀𝑍≀0.64).

We recall that 𝑃(𝑍≀0)=0.5, while 𝑃(0≀𝑍≀0.64)=0.2389 is obtained from the standard normal table. Summing the probabilities, 𝑃(𝑍<0.64)=0.5+0.2389=0.7389.

Converting the probability into a percentage, we get 0.7389Γ—100=73.89%. Rounding to the nearest integer, 74% of blackbirds have masses less than 110 g.

Part 2

Let us find the percentage of blackbirds with masses greater than 124 g. In probability notation, we need to compute 𝑃(𝑋>124). We begin by standardizing the normal distribution: 𝑃(𝑋>124)=𝑃(π‘‹βˆ’πœ‡>21)=π‘ƒο€½π‘‹βˆ’πœ‡πœŽ>2111.

We need to round 2111 to the nearest hundredth, 1.91. Then, the right-hand side of the equation above is equal to 𝑃(𝑍>1.91). We graph the bell curves below to analyze the region {𝑍>1.91}.

Then, we get 𝑃(𝑍>1.91)=𝑃(𝑍β‰₯0)βˆ’π‘ƒ(0≀𝑍≀1.91).

We know that 𝑃(𝑍β‰₯0)=0.5, and we use the standard normal table to obtain 𝑃(0≀𝑍≀1.91)=0.4719. Then, 𝑃(𝑍>1.91)=0.5βˆ’0.4719=0.0281.

Converting the probability into a percentage, we get 0.0281Γ—100=2.81%. Rounding to the nearest tenth, 2.8% of blackbirds have masses greater than 124 g.

Part 3

Let us find the percentage of blackbirds with masses between 95 g and 120 g. In probability notation, we need to compute 𝑃(95≀𝑋≀120). We begin by standardizing the normal distribution: 𝑃(95≀𝑋≀120)=𝑃(βˆ’8β‰€π‘‹βˆ’πœ‡β‰€17)=π‘ƒο€½βˆ’811β‰€π‘‹βˆ’πœ‡πœŽβ‰€1711.

We need to round βˆ’811 and 1711 to the nearest hundredth, βˆ’0.73 and 1.55 respectively. Then, the right-hand side of the equation above is equal to 𝑃(βˆ’0.73≀𝑍≀1.55). We use the symmetry of the bell curve to analyze this probability.

Then, we get 𝑃(βˆ’0.73≀𝑍≀1.55)=𝑃(0≀𝑍≀0.73)+𝑃(0≀𝑍≀1.55).

Using the standard normal table, we get 𝑃(0≀𝑍≀0.73)=0.2673 and 𝑃(0≀𝑍≀1.55)=0.4394. Summing the probabilities, 𝑃(βˆ’0.73≀𝑍≀1.55)=0.2673+0.4394=0.7067.

Converting the probability into a percentage, we get 0.7067Γ—100=70.67%. Rounding to the nearest integer, 71% of blackbirds have masses between 95 g and 120 g.

Example 4: Using Normal Distribution Probabilities to Solve a Real-Life Problem

In a school with 1β€Žβ€‰β€Ž000 students, the heights of students are normally distributed with mean 113 cm and standard deviation 5 cm. How many students are shorter than 121 cm?

Answer

Let 𝑋 represent the height of a student. Then, π‘‹βˆΌπ‘ο€Ή113,5ο…οŠ¨. To answer the question, we first need to determine approximately what percentage of the students are shorter than 121 cm. So we compute 𝑃(𝑋≀121). Standardizing the normal distribution, 𝑃(𝑋≀121)=𝑃(π‘‹βˆ’πœ‡β‰€8)=π‘ƒο€½π‘‹βˆ’πœ‡πœŽβ‰€85=𝑃(𝑍≀1.6).

We draw the bell curve to analyze the probability.

This leads to 𝑃(𝑍≀1.6)=𝑃(𝑍≀0)+𝑃(0≀𝑍≀1.6).

By symmetry, 𝑃(𝑍≀0)=0.5. Using the standard normal table, we get 𝑃(0≀𝑍≀1.6)=0.4452. Adding up the probabilities, 𝑃(𝑍≀1.6)=0.5+0.4452=0.9452.

So 𝑃(𝑋≀121)=0.9452, which means that approximately 0.9452Γ—100=94.52% of the students are shorter than 121 cm. Since we have 1β€Žβ€‰β€Ž000 students total, 94.52% of the total students is 94.52%Γ—1000β‰ˆ945.studentsstudents

We have rounded the right-hand side of the equation above to the nearest integer, since the number of students must be an integer.

So, approximately 945 students are shorter than 121 cm.

The two parameters πœ‡ and 𝜎 characterize a normally distributed random variable. If we are given the values of these two parameters, we can standardize the normal distribution and find the probabilities using the standard normal table. Some problems leave one or both of these parameters unknown. In the next two examples, we will consider problems with unknown parameters.

Example 5: Finding the Mean Using Normal Distribution

The heights of a sample of flowers are normally distributed with mean πœ‡ and standard deviation 12 cm. Given that 10.56% of the flowers are shorter than 47 cm, determine πœ‡.

Answer

Let 𝑋 represent the height of a flower. Then, π‘‹βˆΌπ‘ο€Ήπœ‡,12ο…οŠ¨. Since 10.56% of the flowers are shorter than 47 cm, we know that 𝑃(𝑋≀47)=0.1056. We converted the percentage to a decimal number by dividing by 100.

Standardizing the normal distribution, 𝑃(𝑋≀47)=𝑃(π‘‹βˆ’πœ‡β‰€47βˆ’πœ‡)=𝑃𝑍≀47βˆ’πœ‡12.

Let us denote 𝑧=47βˆ’πœ‡12. We are given that 𝑃(𝑍≀𝑧)=0.1056. In other words, 0.1056 is the probability associated with the 𝑧-score given by 𝑧. Since the probability 0.1056 is less than 0.5, 𝑧 must be negative.

We use the pictures below to think through the process.

This leads to the equations 𝑃(𝑍≀𝑧)=𝑃(𝑍β‰₯βˆ’π‘§)=𝑃(𝑍β‰₯0)βˆ’π‘ƒ(0β‰€π‘β‰€βˆ’π‘§).

Since 𝑃(𝑍β‰₯0)=0.5 and 𝑃(𝑍≀𝑧)=0.1056, this means 0.1056=0.5βˆ’π‘ƒ(0β‰€π‘β‰€βˆ’π‘§)βŸΉπ‘ƒ(0β‰€π‘β‰€βˆ’π‘§)=0.5βˆ’0.1056=0.3944.

Using the standard normal table, the value of 𝑧 corresponding to the probability of 0.3944 is 1.25: 𝑃(0β‰€π‘β‰€βˆ’π‘§)=𝑃(0≀𝑍≀1.25), so βˆ’π‘§=1.25, or equivalently 𝑧=βˆ’1.25. Recall that we have defined 𝑧=47βˆ’πœ‡12. Then, 47βˆ’πœ‡12=βˆ’1.25⟹47βˆ’πœ‡=βˆ’15βŸΉπœ‡=62.

So, πœ‡=62cm.

Example 6: Finding the Mean Using Normal Distribution

The heights of a group of students follow a normal distribution with a standard deviation of 20 cm. The probability that a student’s height is less than or equal to 180 cm is equal to the probability that a standard normal variable is less than or equal to 2.2. Find the mean height of the group of students.

Answer

Let 𝑋 represent the height of a student, which is normally distributed with 𝜎=20. Then, π‘‹βˆΌπ‘ο€Ήπœ‡,20ο…οŠ¨. We notice here that the mean, πœ‡, is unknown and the question asks us to find this value.

We are given that 𝑃(𝑋≀180)=𝑃(𝑍≀2.2), and we remember that 𝑍=π‘‹βˆ’πœ‡πœŽ, so 𝑃(𝑋≀180)=π‘ƒο€½π‘‹βˆ’πœ‡πœŽβ‰€2.2=𝑃(π‘‹βˆ’πœ‡β‰€2.2Γ—πœŽ)=𝑃(π‘‹β‰€πœ‡+2.2Γ—πœŽ).

So this means 180=πœ‡+2.2𝜎. Since we know 𝜎=20, 180=πœ‡+2.2Γ—20βŸΉπœ‡=180βˆ’2.2Γ—20=136.

So, the mean height of the group of students is 136 cm.

Key Points

  • For application problems involving the normal distribution, we begin by defining 𝑋 to be the normal variable with mean πœ‡ and standard deviation 𝜎.
  • If the problem provides variance instead of standard deviation, then we should remember to take the square root to obtain the standard deviation 𝜎.
  • Given a normal random variable 𝑋 with mean πœ‡ and standard deviation 𝜎, we can standardize it using the formula 𝑍=π‘‹βˆ’πœ‡πœŽ.
  • Given a value π‘₯ of a random variable, its 𝑧-score is 𝑧=π‘₯βˆ’πœ‡πœŽ. The probability associated with the 𝑧-score is 𝑃(𝑍≀𝑧).

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