Question

Given a random variable ? has population mean ? = 20 and standard deviation ? = 5.

a.) For the sampling distribution of the sample mean ?, for ? = 50 trials what are the mean and standard deviation? ??bar = _____________, ??bar =_____________.

b.) What is the probability that the sample mean will be in the interval 19 < ?bar < 21 ?

?(19 < ? < 21) =_____________. (give answer to at least 4 signifiant figures.)

Answer 1

Solution :

Given that,

mean = = 20

standard deviation = = 5

a.)

n = 50

The sampling distribution of mean and standard deviation is ,

= 20

= / n = 5 / 50 = 0.7071

b.)

= P[(19 - 20) / 0.7071 < ( - ) / < (21 - 20) / 0.7071)]

= P(-1.41 < Z < 1.41)

= P(Z < 1.41) - P(Z < -1.41)

= 0.9207 - 0.0793

= 0.8414

Probability = 0.8414