High School

The heights of kindergarten children are approximately normally distributed with the following parameters:

Mean (\(\mu\)) = 36 inches
Standard Deviation (\(\sigma\)) = 3.7 inches

(a) If an individual kindergarten child is selected at random, what is the probability that he or she has a height between 34.9 and 37.1 inches?

(b) A classroom of 15 of these children is used as a sample. What is the probability that the class mean \(x\) is between 34.9 and 37.1 inches?

(c) If an individual kindergarten child is selected at random, what is the probability that he or she is taller than 37 inches?

(d) A classroom of 15 of these kindergarten children is used as a sample. What is the probability that the class mean \(x\) is greater than 37 inches?

Answer :

To solve these problems, we will use the properties of the normal distribution.

(a) Probability that an individual's height is between 34.9 and 37.1 inches:

First, let's standardize the height values using the Z-score formula:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Where:


  • [tex]X[/tex] is the value for height.

  • [tex]\mu = 36[/tex] inches is the mean.

  • [tex]\sigma = 3.7[/tex] inches is the standard deviation.


Calculate the Z-scores for 34.9 and 37.1 inches:


  • For 34.9 inches:
    [tex]Z_1 = \frac{34.9 - 36}{3.7} \approx -0.297[/tex]


  • For 37.1 inches:
    [tex]Z_2 = \frac{37.1 - 36}{3.7} \approx 0.297[/tex]



Next, we look up these Z-scores in the standard normal distribution table (or use a calculator):


  • ( P(Z < -0.297) \approx 0.383
    P(Z < 0.297) \approx 0.617 \


The probability that a child’s height is between 34.9 and 37.1 inches is:
[tex]P(-0.297 < Z < 0.297) = P(Z < 0.297) - P(Z < -0.297) \approx 0.617 - 0.383 = 0.234[/tex]

(b) Probability that the class mean [tex]\bar{x}[/tex] is between 34.9 and 37.1 inches for a sample of 15 children:

Since we are dealing with a sample, use the standard error (SE) to calculate:

[tex]SE = \frac{\sigma}{\sqrt{n}} = \frac{3.7}{\sqrt{15}} \approx 0.955[/tex]

Calculate the Z-scores for 34.9 and 37.1 inches using the SE:


  • For 34.9 inches:
    [tex]Z_1 = \frac{34.9 - 36}{0.955} \approx -1.151[/tex]


  • For 37.1 inches:
    [tex]Z_2 = \frac{37.1 - 36}{0.955} \approx 1.151[/tex]



Then, find the probabilities from the Z-table:


  • ( P(Z < -1.151) \approx 0.125
    P(Z < 1.151) \approx 0.875 \


Thus, the probability is:
[tex]P(-1.151 < Z < 1.151) = P(Z < 1.151) - P(Z < -1.151) \approx 0.875 - 0.125 = 0.750[/tex]

(c) Probability that a randomly selected child is taller than 37 inches:

Calculate the Z-score for 37 inches:

[tex]Z = \frac{37 - 36}{3.7} \approx 0.270[/tex]

Find [tex]P(Z > 0.270)[/tex]:

[tex]P(Z > 0.270) = 1 - P(Z < 0.270) \approx 1 - 0.606 = 0.394[/tex]

(d) Probability that the class mean [tex]\bar{x}[/tex] is greater than 37 inches for a sample of 15 children:

Use the standard error to calculate the Z-score for 37 inches:

[tex]Z = \frac{37 - 36}{0.955} \approx 1.047[/tex]

Find [tex]P(Z > 1.047)[/tex]:

[tex]P(Z > 1.047) = 1 - P(Z < 1.047) \approx 1 - 0.852 = 0.148[/tex]

This means there is approximately a 14.8% probability that the average height of a sample class of 15 children is greater than 37 inches.

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