Question

A population of 1,000 students spends an average of $10.50 a day on dinner. The standard deviation of the expenditure is $3. A simple random sample of 64 students is taken.

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

b. What is the probability that these 64 students will spend an average of more than $11 per person?

c. What is the probability that these 64 students will spend an average between $10 and $11 per person?

Answer 1

Solution :

Given that ,

mean = = 10.50

standard deviation = = 3

a) n = 64

=   = 10.50

= / n = 3 / 64 = 0.375

b) P( > 11) = 1 - P( < 11)

= 1 - P[( - ) / < (11 - 10.50) / 0.375]

= 1 - P(z < 1.33)

Using z table,    

= 1 - 0.9082

= 0.0918

c) P(10 < < 11)  

= P[(10 - 10.50) /0.375 < ( - ) / < (11 - 10.50) / 0.375)]

= P(-1.33 < Z < 1.33)

= P(Z < 1.33) - P(Z < -1.33)

Using z table,  

= 0.9082 - 0.0918  

= 0.8164