Question

The mean weight of an adult is 71

71

kilograms with a variance of 100

100

.

If 181

181

adults are randomly selected, what is the probability that the sample mean would differ from the population mean by greater than 1.2

1.2

kilograms? Round your answer to four decimal places.

Answer 1

Solution :

Given that,

mean = = 71

variance= 100

standard deviation = = 10

n = 181

= 71

= / n = 10 / 181

1 - P(69.8 < < 72.2) = 1 - P((69.8 - 71) / 10 / 181 <( - ) / < (72.2 - 71) / 10 / 181))

= 1 - P(1.61 < Z < 1.61)

= 1 - [P(Z < 1.61) - P(Z < -1.61) ] Using z table,

= 1 - [0.9463 - 0.0537]

= 1- 0.8926

= 0.1074

Probability = 0.1074