Question

(8) A normal population has a mean of 50 and unknown variance. (a) A random sample of n = 16 is selected from this population and the sample results are x = 52 and s = 8. What is the probability of observing a sample average as large as 52 (or larger) if the mean is actually 50?

please explain the steps clearly

Answer 1

Answer 8 a

Given, a normal population has mean = 50 and unknown variance

Let x be a random variable representing the above population

x follows N(50, V) (V – Unknown Variance)

Since, x follows Normal Distribution

also follows Normal Distribution with mean = 52 (given) and standard deviation, s = 8 (given)

Therefore, follows N(52, 8)

The required probability, P[ ≥ 52] = P[( - 50)/8 ≥ (52 - 50)/8] (Standardizing…)

                                             = P[Z ≥ 0.25] (where Z follows Standard Normal Distribution)

                                            = 1 – P[Z < 0.25] = 1 – 0.5987 (value taken from Standard Normal Table)

                                           = 0.4013 (up to 4 decimal places)

Therefore, Probability of observing a sample average as large as 52 (or larger) is 0.4013