Question

A population has a normal distribution with a mean of 51.4 and a standard deviation of 8.4. Assuming n/N is less than or equal to 0.05, the probability, rounded to four decimal places, that the sample mean of a sample size of 18 elements selected from this population will be more than 51.15 is?

Answer 1

Solution :

Given that ,

mean = = 51.4

standard deviation = = 8.4

n = 18

= = 51.4 and

= / n = 8.4 / 18 = 1.9799

P( > 51.15) = 1 - P( < 51.15)

= 1 - P(( - ) / < (51.15 - 51.4) / 1.9799)

= 1 - P(z < -0.13)

= 1 - 0.4483

= 0.5517

Probability = 0.5517