Question

At the end of the Halloween Festival the organizers estimated that a family of participants spent...

At the end of the Halloween Festival the organizers estimated that a family of participants spent in average $45.00 with a standard deviation of $10.00.

If 49 participants (49 = size of the sample) are selected randomly, what’s the likelihood that their mean spent amount will be within $4 of the population mean? (mean +/- 4)

Math   Statistics-And-Probability   
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Answer #1

We have

= 45 and = 10

If n = 49

Xbar - = -4 and xbar + = 4

We asked P( -4 < xbar -  < 4 )

P ( -4/(/√n) < Z < 4/(/√n) ) = P(-2.80 < Z < 2.80)

P( -4 < xbar - < 4 ) = P ( Z < 2.80) - P( Z < -2.80)

= 0.9974 - 0.0026

P( -4 < xbar - < 4 ) = 0.9948

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