Answer to Question #279118 in Statistics and Probability for Aly

Question

The heights of preschool children as reported by the Department of Statistics is approximately normally distributed about a mean of 39 inch and a standard deviation of 2 inch.  

1. What is the probability that the mean value is more than 40.0 inch?
2. A random sample of size 25 is taken. What is the probability that the mean 𝑋̅ is more than 40.0 inch?
3. Why is the answer for (b) is smaller than (a)?
4. Within what limits does the middle 90% of the sampling distribution of sample means for samples of size 100 fall?

Answer 1

1.

$P(X>40) = 1 -P(X<40) \\ =1 -P(Z< \frac{40-39}{2}) \\ = 1- P(Z<0.5) \\ = 1- 0.6914 \\ =0.3086$

2.

$n=25 \\ P(\bar{X} > 40) = 1 -P(\bar{X} < 40) \\ =1 -P(Z < \frac{40-39}{2 / \sqrt{25}}) \\ = 1 -P(Z < 2.5) \\ = 1 -0.9937 \\ = 0.0063$

3.

The answer for (b) is smaller than (a) because the (b) answer represents the probability of the sample mean with limited number of children, while (a) answer shows the population probability.

4.

$n=100 \\ CI = (\mu - \frac{Z_c \times \sigma}{\sqrt{n}}, \mu + \frac{Z_c \times \sigma}{\sqrt{n}}) \\ Z_c(90\%)=1.645 \\ CI = (39 - \frac{1.645 \times 2}{\sqrt{100}}, 39 + \frac{1.645 \times 2}{\sqrt{100}}) \\ =(39 -0.329, 39 + 0.329) \\ Lower \; limit = 38.671 \\ Upper \; limit = 39.329$