Question

A population of values has a normal distribution with μ = 8.2 and σ = 30.2 . You intend to draw a random sample of size n = 28 . Find the probability that a single randomly selected value is greater than -0.9. P(X > -0.9) = Find the probability that a sample of size n = 28 is randomly selected with a mean greater than -0.9. P(M > -0.9) = Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

Solution :

Given that ,

mean = = 8.2

standard deviation = = 30.2

n = 28

M = 8.2 and

M = / n = 30.2 / 28 = 5.7073

P(M > -0.9) = 1 - P(M < -0.9)

= 1 - P(( - ) / < (-0.9 - 8.2) / 5.7073)

= 1 - P(z < -1.594)

= 1 - 0.0555

= 0.9445

P(M > -0.9) = 0.9445

Probability = 0.9445