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?

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

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