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sergij07 [2.7K]
3 years ago
6

Suppose a large shipment of telephones contained 21% defectives. If a sample of size 498 is selected, what is the probability th

at the sample proportion will differ from the population proportion by less than 3%
Mathematics
1 answer:
grandymaker [24]3 years ago
6 0

Answer:

P(-3\% < x < 3\%) = 0.901

Step-by-step explanation:

Given

p = 21\%

n = 498

Required

P(-3\% < x < 3\%)

First, we calculate the z score

z = \sqrt{p * (1 - p)/n}

z = \sqrt{21\% * (1 - 21\%)/498}

z = \sqrt{21\% * (79\%)/498}

z = \sqrt{0.1659/498}

z = \sqrt{0.000333}

z = 0.0182

So:

P(-3\% < x < 3\%) = P(-3\%/0.0182 < z

P(-3\% < x < 3\%) = P(1.648 < z

From z probability, we have:

P(-3\% < x < 3\%) = 0.901

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Step by step solution :

Step  1  :

Equation at the end of step  1  :

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Step  2  :

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The middle term is,  -5n  its coefficient is  -5 .

The last term, "the constant", is  -6  

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Step-2 : Find two factors of  -24  whose sum equals the coefficient of the middle term, which is   -5 .

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Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -8  and  3  

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Step-4 : Add up the first 2 terms, pulling out like factors :

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             Add up the last 2 terms, pulling out common factors :

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