Next 5 questions are based on the following information: The mean annual cost of

Question

Next 5 questions are based on the following information: The mean annual cost of full-coverage auto insurance is µ = $1,820. Assume the population standard deviation is = $410. 5 The standard error of the sampling distribution of the mean of random samples of size n = 64 automobile insurance policies is, a 53.25 b 52.50 c 51.25 d 49.75 6 In the previous question, the fraction of sample means the fall within the interval ($1,740, $1,900) is, a 0.8664 b 0.8812 c 0.9010 d 0.9198 7 The margin of error for the middle interval which captures 95% of the means from samples of size n = 64 is, a 65.60 b 84.05 c 96.55 d 100.45 8 In the previous question, the middle interval that captures 95% of all sample means from samples of size n = 64 is, a 1719.55 1920.45 b 1735.95 1904.05 c 1754.40 1885.60 d 1763.63 1876.38 9 To achieve a 95% margin or error of ±$50 for the sampling distribution, the minimum sample size is, a 226 b 245 c 259 d 268 Next 5 questions are based on the following information: The mean annual cost of full-coverage auto insurance is µ = $1,820. Assume the population standard deviation is = $410. 5 The standard error of the sampling distribution of the mean of random samples of size n = 64 automobile insurance policies is, a 53.25 b 52.50 c 51.25 d 49.75 6 In the previous question, the fraction of sample means the fall within the interval ($1,740, $1,900) is, a 0.8664 b 0.8812 c 0.9010 d 0.9198 7 The margin of error for the middle interval which captures 95% of the means from samples of size n = 64 is, a 65.60 b 84.05 c 96.55 d 100.45 8 In the previous question, the middle interval that captures 95% of all sample means from samples of size n = 64 is, a 1719.55 1920.45 b 1735.95 1904.05 c 1754.40 1885.60 d 1763.63 1876.38 9 To achieve a 95% margin or error of ±$50 for the sampling distribution, the minimum sample size is, a 226 b 245 c 259 d 268

Explanation / Answer

5) S = 410 , n = 64

Standard Error = S / sqrt(n)

= 410 / sqrt(64)

= 51.25

6)

here, X1 = 1740 , X2 = 1900 , mean = 1820 , SE = 51.25

By normal distribution formula,

Z = ( X1 - mean) / SE

= ( 1740 - 1820 ) /51.25

= -1.560

Z = ( X2 - mean ) /SE

= ( 1900 - 1820) / 51.25

= 1.560

Now, we need to find P( Z < 1.560) and P(Z < -1.560) , By using z standard right tail we get,

X = 1 - P( Z < 1.560) - P(Z < -1.560)

= 1 - .0594 - .0594

X = .8812

7)

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