Finding probabilities for normal distribution

MTH 245 Lesson 15 Notes Finding Probabilities for the Normal Distribution

To calculate normal distribution probabilities using StatCrunch:

1. Select Stats  Calculators  Normal.

2. Click "Standard" for one-sided inequalities or "Between" for two- sided.

3. Input 𝜇𝜇 and 𝜎𝜎. 4. Adjust the operator inside the parentheses on the lowest line.

5. Leave the field to the right of the equal sign blank, then input the appropriate value(s) inside the parentheses.

6. Click "Compute!". Example 1: For a random variable 𝑥𝑥 with mean 𝜇𝜇 = 100 and standard deviation 𝜎𝜎 = 15, find the following:

a. 𝑃𝑃(𝑥𝑥 ≤ 80) 0.091

b. 𝑃𝑃(𝑥𝑥 ≥ 123) 0.063

c. 𝑃𝑃(70 ≤ 𝑥𝑥 ≤ 112) 0.765

To find the values associated with a known probability, follow the same steps above, except in Step 5, leave the field(s) inside the parentheses blank, and input the probability in the field to the right of the equal sign.

Example 2: For the normal random variable 𝑥𝑥 in Example 1, find the following:

a. 𝑥𝑥0.975 (the 97.5th percentile of 𝑥𝑥) 129.4

b. 𝑥𝑥0.05 (the 5th percentile of 𝑥𝑥) 75.3

c. The value of 𝑥𝑥 such that the area to the right of 𝑥𝑥 is 0.10 119.2

d. The two values of 𝑥𝑥 such that the area between them is 0.75

82.7, 117.3

Example 3: A certain airline requires that its pilots be between 62 and 73 inches tall. Suppose men's heights are normally distributed with 𝜇𝜇 = 69.5 inches and 𝜎𝜎 = 2.4 inches.

a. What percentage of men meet the qualification (i.e., what is the probability that a randomly selected man is between 62 and 73 inches tall)? 92.7%

b. What is the 95th percentile of men's heights? 73.4 inches

Example 4: Suppose scores on a standardized test are normally distributed with 𝜇𝜇 = 450 and 𝜎𝜎 = 125.

a. What is the probability that a random test-taker will score above 700? 2.3%

b. Find the score that separates the top five percent of test takers from

the bottom 95 percent (round to the nearest integer). 656

Example 5. After 1964, quarters were manufactured so their weights have a mean of 5.67 g and a standard deviation of 0.06 g. Some vending machines are designed so that the weights of quarters that are accepted can be adjusted. If many counterfeit coins are found, the weight can be adjusted so that most of the counterfeits are rejected, although this also means some legal quarters will also be rejected.

a. If a vending machine is adjusted to accept quarters with weights

between 5.60 g and 5.74 g, what percentage of legal quarters will be rejected? 24% will be rejected

b. If the machine is adjusted to accept all legal quarters except those

with weights in the top 2.5% and bottom 2.5%, what are the limits of weights that will be accepted? 5.55 g, 5.79 g

Significance and the Normal Distribution

For any random variable 𝑥𝑥, the critical values are those values of 𝑥𝑥 that separate significant values from those that are not significant. If the distribution of 𝑥𝑥 is symmetric, these two values will be the negative of each other. To determine which values are significant, we need to first define the significance level. This value, which we will denote 𝛼𝛼, is the proportion of possible values of 𝑥𝑥 that are either significantly low or significantly high. The remaining area under the bell curve, 1 − 𝛼𝛼, is the proportion of possible values of 𝑥𝑥 that are not significant. The graphic below illustrates the relationship between the regions (significant values shaded, non-significant values in unshaded):

To find the critical values with StatCrunch, use the normal distribution calculator to find the two 𝑥𝑥 values for which the area between them equals 1 − 𝛼𝛼. We saw how to do this in Example 2d above.

Example 6: Find the standard normal (𝜇𝜇 = 0, 𝜎𝜎 = 1) critical values for the following values of 𝛼𝛼 (round your answers to two decimal places):

a. 𝛼𝛼 = 0.05 The two values where 1 − 𝛼𝛼 = 0.95 of the area lies between them are −1.960, 1.960.

b. 𝛼𝛼 = 0.01 The two values where 1 − 𝛼𝛼 = 0.99 of the area lies between them are −2.575, 2.575.

c. 𝛼𝛼 = 0.10 The two values where 1 − 𝛼𝛼 = 0.90 of the area lies between them are −1.645, 1.645.

Example 7: For a normal distribution with 𝜇𝜇 = 100, 𝜎𝜎 = 16, find the value of 𝛼𝛼 that corresponds to the following sets of critical values (round your answers to three decimal places):

a. 73.6823, 126.3177 1 − 𝛼𝛼 = 0.900, so 𝛼𝛼 = 0.010.

b. 62.7784, 137.2216 1 − 𝛼𝛼 = 0.980, so 𝛼𝛼 = 0.020.

c. 55.0875, 144.9125 1 − 𝛼𝛼 = 0.995, so 𝛼𝛼 = 0.005.

Example 8: Assume that a certain type of precision-machined bolt is has a length that is approximately normally distributed with a mean of 4.000 cm and standard deviation of 0.010 cm. The manufacturer has determined that approximately 5% of the bolts will be rejected during post-production inspection because they are too long, and another 5% will be rejected because they are too short. Find the critical values – that is, the two measurements (in cm) that can be used as cutoff values separating the accepted bolts from the rejects. Round your answers to three decimal places.

The area between the critical values must equal 1 − (0.05 + 0.05) = 0.90. Therefore, the critical values are 3.984 and 4.016. Bolts with lengths between 3.984 cm and 4.016 cm will be accepted, while bolts with lengths less than 3.984 cm or greater than 4.016 cm will be rejected.