Question

A population's distribution is normal with a mean of 18 and a standard deviation of 4. A sample of 20 observations is selected and a sample mean computed. What is the probability that the sample mean is more than 19.5? Express your solution as the decimal equivalent of a probability rounded to four decimal places.

Answer 1

Let X be population

X follows normal distribution with mean : of 18 and standard deviation : 4

By Central limit theorem, if X follows a normal distribution with mean and standard deviation ; Then a sample distribution of sample mean ( with sample size : n;) follows normal distribution with mean and standard deviation

Therefore ,

Sample mean: with sample size: 20 follows normal distribution   and standard deviation

Probability that the sample mean is more than 19.5 = P( > 19.5) = 1-P(19.5)

z-score for 19.5 = (19.5 - 18)/0.8944 = 1.68

From standard normal tables , P(Z1.68) = 0.9535

P(19.5) =P(Z1.68) = 0.9535

P( > 19.5) = 1-P(19.5) = 1-0.9535 = 0.0465

Probability that the sample mean is more than 19.5 = 0.0465