Question

A random sample of 32 is obtained from a population with (mean = 32) (standard deviation= 10)

A) Describe the sampling distribution of the sample mean.

B) Find the sample that has the 60.5th percentile

C) Find the probability of the mean is between 29 and 36

Answer 1

Solution :

Given that,

mean = = 32

standard deviation = = 10

n = 32

A)

= 32

= / n = 10 / 32 = 1.7678

B)

Using standard normal table ,

P(Z < z) = 60.5%

P(Z < 0.27) = 0.605

z = 0.27

Using z-score formula,

Using z-score formula,  

= z * +   = 0.27 * 1.7678 + 32 = 32.48

The 60.5th percentile is 32.48

C)

= P[(29 - 32) / 1.7678 < ( - ) / < (36 - 32) / 1.7678)]

= P(-1.70 < Z < 2.26)

= P(Z < 2.26) - P(Z < -1.70)

= 0.9881 - 0.0446

= 0.9435

Probability = 0.9435