Question

Students of a large university spend an average of $5 a day on lunch. The standard deviation of the expenditure is $3. A simple random sample of 36 students is taken.

What are the expected value, standard deviation, and shape of the sampling distribution of the sample mean?

What is the probability that the sample mean will be at least $4?

What is the probability that the sample mean will be at least $5.90?

Answer 1

Solution :

Given that ,

mean = = 5

standard deviation = = 3

n = 36

Expected value = = 5 and

standard deviation = = / n = 3 / 36 = 3 / 6 = 0.5

The shape of the sampling distribution of the sample mean is approximately normal .

(a)

The probability that the sample mean will be at least $4 is,

P( 4) = 1 - P(​ 4)

P( 4) = 1 - P(( - ) / (4 - 5) / 0.5)

P( 4) =   1 - P(z < -2)

P( 4) = 1 -  0.0228 = 0.9772

Using standard normal table

Probability = 0.9772

b)  The probability that the sample mean will be at least $5.90 is,

P( 5.90) = 1 - P(​ 5.90)

P( 5.90) = 1 - P(( - ) / (5.90 - 5) / 0.5)

P( 5.90) = 1 - P(z < 1.8)

P( 5.90) = 1 - 0.9641 = 0.0359

Probability = 0.0359