Answer:
Given that the population of men has normally distributed weights, with a mean of 173 lb and a standard deviation of 30 lb, find the probability that
a. if 1 man is randomly selected, his weight is greater than 180 lb.
b. if 36 different men are randomly selected, their mean weight is greater that 180 lb.
Solution: a) z = (x - μ)/ σ = (180-173)/30 = 0.23
For normal distribution P(Z>0.23) = 0.4090
b) σ x̄ = σ/√n = 20/√ 36 = 5
z= (180-173)/5 = 1.40
P(Z>1.4) = 0.0808
a. if 1 man is randomly selected, his weight is greater than 180 lb.
b. if 36 different men are randomly selected, their mean weight is greater that 180 lb.
Solution: a) z = (x - μ)/ σ = (180-173)/30 = 0.23
For normal distribution P(Z>0.23) = 0.4090
b) σ x̄ = σ/√n = 20/√ 36 = 5
z= (180-173)/5 = 1.40
P(Z>1.4) = 0.0808
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