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Ouesion 1 Not yet answered Points out of 2.00 P Flag auestion In engineering and product design, it is important to consider
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Answer #1

Solution:

We are given that: Weight of adult makes in the US has a mean weight of 197 pounds and standard deviation of 32 pounds.

That is: Mean = mu = 197 and Standard Deviation = sigma = 32

n = sample size = 50

We have to find the probability that: the average weight of these 50 adult males is over 200 pounds.

That is we have to find:

P( ar{X}>200)=.........?

z=rac{ar{x}-mu }{sigma /sqrt{n}}

z=rac{200-197 }{32 /sqrt{50}}

z=rac{ 3 }{32 /7.071068 }

z=rac{ 3 }{4.52548 }

z=0.66

Thus we get:

P( ar{X}>200)=P(Z > 0.66)

P( ar{X}>200)=1 - P(Z < 0.66)

To get P( Z < 0.66) , look in z table for z = 0.6 and 0.06 and find area.

01 02 03 .04 05 06 07 08 0.0 5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 0.1 .5398 5438 .5478 5517 .5557 .5 0.

P( Z < 0.66) = 0.7454

Thus

P( ar{X}>200)=1 - P(Z < 0.66)

P( ar{X}>200)=1 - 0.7454

P( ar{X}>200)=0.2546

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