Question

A normal distribution has a mean of 30 and a variance of 5.Find N such that the probability that the mean of N observations exceeds 30.5 is 1%.​

Answer 1

Answer:

109

Step-by-step explanation:

Use a chart or calculator to find the z-score corresponding to a probability of 1%.

P(Z > z) = 0.01

P(Z < z) = 0.99

z = 2.33

Now find the sample standard deviation.

z = (x − μ) / s

2.33 = (30.5 − 30) / s

s = 0.215

Now find the sample size.

s = σ / √n

s² = σ² / n

0.215² = 5 / n

n = 109