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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