Answer:
0.1994 is the required probability.
Step-by-step 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:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Standard error due to sampling =
[tex]=\dfrac{\sigma}{\sqrt{n}} = \dfrac{5.3}{\sqrt{20}} = 1.1851[/tex]
P(sample of 20 boxers is more than 167 pounds)
[tex]P( x > 167) = P( z > \displaystyle\frac{167 - 166}{1.1851}) = P(z > 0.8438)[/tex]
[tex]= 1 - P(z \leq 0.8438)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 167) = 1 - 0.8006= 0.1994 = 19.94\%[/tex]
0.1994 is the probability that the mean weight of a random sample of 20 boxers is more than 167 pounds
