The mean life of a television set is 96 months with a variance of 225. If a sample of 76 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by greater than 1.3 months?Round your answer to four decimal places.

0.4496

Step-by-step explanation:

Mean u = 96

Sd² = 225

Sd = √225

= 15

n = 76

Bar x is the mean

Bar x follows an approximately normal distribution

sd/√n = 15/√76

= 1.720618

The probability that bar x would differ from mean by less than -1.3

P(-1.3<z<1.3)

P(-1.3/1.720618 <z <1.3/1.720618)

= P(-0.756 < z < 0.756)

= P(z<0.756)-p(z<-0.756)

= 0.7752-0.2242

= 0.5504

To get the probability

1-0.5504

= 0.4496

nlgqer

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