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katen-ka-za [31]
3 years ago
11

In a large population, 76% of the households have cable tv. A simple random sample of 225 households is to be contacted and the

sample proportion computed. What is the probability that the sampling distribution of sample porportions is less than 82%?
Mathematics
1 answer:
serious [3.7K]3 years ago
6 0

Answer:

0.017559

Step-by-step explanation:

Data provided:

Probability of Households Having cable TV, p₀ = 76% = 0.76

Therefore,

The probability that the Households not having cable TV = 1 - 0.76 = 0.24

Sample size, n = 225 households

sample proportions is less than 82% i.e p = 0.82

Now,

The standard error, SE = \sqrt{\frac{p_0(1-p_0)}{n}}

or

SE = \sqrt{\frac{0.76(1-0.76)}{225}}

or

SE = 0.02847

and,

Z=\frac{p-p_0}{SE}

or

Z=\frac{0.82-0.76}{0.02847}

or

Z = 2.107

therefore,

P(sample porportions < 0.82) = P(Z < 2.107)

now from the p value from the Z table

we get

P(sample porportions < 0.82) =  0.017559

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The option B is the correct option.

Given information-

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<h3>Power rule of exponents</h3>

Power rule of exponents says that the power of the power of a exponent is equal to the multiplying both the powers.

Suppose<em> </em><em>x</em> is a number with power <em>a. </em>The power is power of the power of number <em>x. </em>Then by the power rule of the exponents,

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The option B is the correct option.

Learn more about the power rule of exponents here;

brainly.com/question/819893

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