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Sagot :
If 17% of men are bald, then the probability that at most 140 in a random sample of 900 men are bald, is 0.1335.
In the given question, if 17% of men are bald, then we have to find the probability that at most 140 in a random sample of 900 men are bald.
As given that; 17% of men are bald.
So,p=0.17
Let X=Number of men that are bald
Sample size, n=900
Here X follows binomial distribution with parameters n= 900 and p =0.17
Since np≥5 and n(1-p)≥5, We can use normal approximation to the binomial with continuity correction.
So,Binomial can be approximated to normal with;
mean, μ=np
μ = 900*0.17
μ = 153
Standard deviation, σ = √np(1-p)
σ = √900*0.17*(1-0.17)
σ = √153*0.83
σ = √126.99
σ = 11.269
So, X→Normal (μ = 153, σ = 11.269)
Then, X= (X-μ)/σ
X=(X-153)/11.269
We need to find the probability that at most 140 in a random sample of 900 men are bald i.e. we need to find P(X≤140)
P(X≤140) =P(X<140+0.5) [using continuity correction factor]
P(X≤140) =P(X<140.5)
P(X≤140) =P(z<(X-153)/11.269)
P(X≤140) =P(z<(140.5-153)/11.269)
P(X≤140) =P(Z<-1.11) [z score rounded to 2 decimal places]
P(X≤140) =0.1335
Hence, if 17% of men are bald, then the probability that at most 140 in a random sample of 900 men are bald, is 0.1335.
To learn more about probability link is here
brainly.com/question/29657446
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