Question

1) Students of a large university spend an average of $10 a day on lunch. The standard deviation of the expenditure is $3. A simple random sample of 36 students is taken.

1. What are the expected value, standard deviation, and shape of the sampling distribution of the sample mean?
2. What is the probability that the sample mean will be at least $9.00?
3. What is the probability that the sample mean will be greater than $10.50?

Answer 1

Solution :

a.

expected Value = E( X ) = μ = $10

standard deviation = σ / sqrt ( n ) = $3 / sqrt(36) = 0.50

shape of the sampling distribution of the sample mean : Approximately normal with mean $10 and standard deviation $0.50

b.

What is the probability that the sample mean will be at least $9.00?

P( x bar >= $9.00) = P( Z >= (9-10)/0.5)

= P( Z > -2 )

= 0.9772

c.

What is the probability that the sample mean will be greater than $10.50?

P( x bar > $10.50) = P( Z >= (10.50-10)/0.5)

= P( Z > 1 )

= 0.1587