7.1 Lecture Notes (2)pdf Practice 7.1 The Central l name-firstname-tec × 3view s

Question

7.1 Lecture Notes (2)pdf Practice 7.1 The Central l name-firstname-tec × 3view start%3D2018-... assignments/95561 ?return-to: https%3A%2F%2Frsc Due in 5 hours, 3 minutes. Due Thu 03/22/2018 11:59 pm A manufacturer knows that their items have a normally distributed lifespan, with a mean of 11.6 years, and standared devlation of 3.1 years If you randomly purchase 23 items, what is the probability that their mean life will be longer than 12 years? Box 1: Enter your answer as an integer or decimal number. Examples: 3,-4, 5.5172 Enter DNE for Does Not Exist, oo for Infinity Points possible: 2 This is atcempr 1 of 3 Submit 42

Explanation / Answer

X : Lifespan of item

X ~ N ( 11.6 , 3.12)

n =23

Then Xbar ~ N ( 11.6 , 3.12/23)

E(Xbar) = 11.6 and S.D(Xbar) = sqrt(3.12/23) = 0.6463)

By Central limit theorem

Z = (Xbar-E(Xbar)) / S.D(xbar) ~ N(0,1)

P(mean life of item will be longer than 12 years ) = P(xbar >12)

=P((Xbar-E(Xbar)) / S.D(xbar) > ( 12-11.6) / 0.6463)

= P(Z > 0.6189)

From normal probability table

P(Z> 0.6189) = 0.2680

Probability of mean life of 23 items will be longer than 12 years is 0.6280.

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