Question

Suppose the weights of the Boxers at this club are Normally distributed with a mean of 166 pounds and a standard deviation of 5.3 pounds. What is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds?

Answer 1

0.1994 is the required probability.

Explanation:

We are given the following information in the question:

Mean, μ = 166 pounds

Standard Deviation, σ = 5.3 pounds

Sample size, n = 20

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

Standard error due to sampling =

`=(\sigma)/(√(n)) = (5.3)/(√(20)) = 1.1851`

P(sample of 20 boxers is more than 167 pounds)

`P( x > 167) = P( z > \displaystyle(167 - 166)/(1.1851)) = P(z > 0.8438)`

`= 1 - P(z \leq 0.8438)`

Calculation the value from standard normal z table, we have,

`P(x > 167) = 1 - 0.8006= 0.1994 = 19.94\%`

0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds