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A population of values has a normal distribution with u = 232 and o = 21.4. A random sample of size n = 85is drawn. Find the probability that a sample of size n = 85 is randomly selected with a mean greater than 235.9.Round your answer to four decimal places.PM > 235.9)

User FatCop
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1 Answer

1 vote

Given:


\mu=232,\text{ }\sigma=21.4,\text{ n=85 and M=235.9}

The z-score value is


z=\frac{M-\mu}{\frac{\sigma}{\sqrt[]{n}}}


\text{ Substitue }\mu=232,\text{ }\sigma=21.4,\text{ n=85 and M=235.9, we get}


z=\frac{235.9-232}{\frac{21.4}{\sqrt[]{85}}}


z=\frac{\sqrt[]{85}(235.9-232)}{21.4}


z=1.68019

P value from the z table is


P(-235.9M<235.9)=0.45354
P(M<235.9)-0.5=0.45354


P(M<235.9)=0.95354

We know that


P(M>235.9)=1-P(M<235.9)


P(M>235.9)=1-0.95354


P(M>235.9)=0.046459

Hence the answer is


P(M>235.9)=0.046459

User Jordanna
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