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A population of values has a normal distribution with j = 11.8 and o = 70.8. A random sample of sizen = 170 is drawn.a. Find the probability that a single randomly selected value is between - 1.8 and 28.6. Round your answer tofour decimal places,PC - 1.8 < X < 28.6) =b. Find the probability that a sample of size n = 170 is randomly selected with a mean between -1.8 and28.6. Round your answer to four decimal places.PC - 1.8 < M < 28.6)

A population of values has a normal distribution with j = 11.8 and o = 70.8. A random-example-1
User SGJ
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11 votes

Solution

(a) mean = 11.8

standard deviation = 70.8

P(-1.8 < x < 28.6)


\begin{gathered} z=(x-\mu)/(\sigma)=(-1.8-11.8)/(70.8) \\ z=-(13.6)/(70.8) \\ z=-0.192 \end{gathered}
\begin{gathered} z=(x-\mu)/(\sigma) \\ z=(28.6-11.8)/(70.8)=(16.8)/(70.8)=0.237 \end{gathered}

therefore the z score for p(-1.8 [tex]\begin{gathered} Pr(-0.192Therefore probability that a single randomly selected value between - 1.8 and 28.6 = 0.1699 (4d.p)

User Garpunkal
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