Question

Based on the U.S. Department of Agriculture’s Center for Nutrition Policy and Promotion study, the average cost of raising a child born in 2008 for families in the $56,870 to $98,470 income group is $221,190 with a standard deviation of $30,000. Suppose a random sample of 36 such families was obtained.

The expected value of the average expenditures for the random sample is ______.

The standard error of the average expenditures for the random sample is ______.

The probability that the sample mean expenditure is less than $220,190 is ______.

The probability that the sample mean expenditure is greater than $224,190 is ______.

Answer 1

The expected value of the average expenditures for the random sample is E(X) = 221190

The standard error of the average expenditures for the random sample is sd/sqrt(n) = 30000/sqrt(36)

= 5000

The probability that the sample mean expenditure is less than $220,190 is  

Z =- (Xbar - 221190)/5000

P(Xbar < 220190)

= P (Z<−0.2)=0.4207

The probability that the sample mean expenditure is greater than $224,190

P(Xbar > 224190)

= P (Z>0.6)=0.2743

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