Question

Suppose that the average time it takes a couple to decorate a gingerbread house follows a normal distribution with a population mean of 42.50 minutes and a population variance of 169 minutes-squared. Suppose that a sample of 24 couples has been randomly selected. What is the probability that the sample mean is between 47.75 minutes and 49.25 minutes? Express your solution as the decimal equivalent of a probability rounded to four decimal places. Your Answer: Answer

Answer 1

Values are given as ,

Mean = 42.50

$\sigma = \sqrt{169}$ = 13

$\sigma_x = (\sigma/\sqrt{n}) = (13/\sqrt{24})$ = 2.6536

We need to find P(47.75 < x < 49.25)

Using below formula
$P[(x_1 - \mu)/\sigma_x < z < (x_2 - \mu)/\sigma_x]$

Put the values in the formula,
P[(47.75 - 42.5)/2.6536 < z < (49.25 - 42.5)/2.6536]
= (1.98 < z < 2.54)
z scores are (1.98, 2.54)

Using below formula,
P(a < z < b) = P(z < b) - P(z < a)

Put the values in above formula,
P(1.98 < z < 2.54) = P(z < 2.54) - P(z < 1.98)

Using z table, we get,
P(z < 2.54) = 0.99446
P(z < 1.98) = 0.97615
So, P(1.98 < z < 2.54) = 0.99446 - 0.97615
= 0.0183

Hence, P(1.98 < z < 2.54) = 0.0183

Required answer : [0.0183]