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?

Answer: 10.5     0.363   normal

b. What is the probability that these 64 students will spend a combined total between $10.99 and $11.38?

c. What is the probability that these 64 students will spend a combined total of more than 11.18?

Answer 1

Solution :

Given that,

= 10.5

= 0.363

b)  P(10.99 < < 11.38)  

= P[(10.99 - 10.5) /0.363 < ( - ) / < (11.38 - 10.5) / 0.363)]

= P(1.35 < Z < 2.42)

= P(Z < 2.42) - P(Z < 1.35)

Using z table,  

= 0.9922 - 0.9115

= 0.0807

c) P( > 11.18) = 1 - P( < 11.18)

= 1 - P[( - ) / < (11.18 - 10.5) /0.363 ]

= 1 - P(z < 2.42)

Using z table,    

= 1 - 0.9922

= 0.0078