Chapter 7 Q19

a. The population standard deviation for this problem is 4000 (see page 278)

We can use Excel to find first the probability that the sample is in the area UP TO 51800 + 500 = 52300 and then UP TO 51800 – 500 = 51300. First find the standard error for a sample size of 60. This is 516.4. Then use normdist

=normdist(52300,51800,516.4,true) = 0.83

=normdist(51300,51800,516.4,true) = 0.17

then subtract to get 0.66

b. For (b), recalculate the standard error with the sample size of 120.

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