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.