The mean annual cost of full-coverage auto insurance is mu = $1, 820. Assume the

Question

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

Explanation / Answer

Q5.
Standard Error= sd/ Sqrt(n)
Where,
sd = Standard Deviation
n = Sample Size
Standard deviation( sd )=410
Sample Size(n)=64
Standard Error = ( 410/ Sqrt ( 64) )
= 51.25
Q7.
Margin of Error = Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
Mean(x)=52.4
Standard deviation( sd )=410
Sample Size(n)=64
Margin of Error = Z a/2 * 410/ Sqrt ( 64)
= 1.96 * (51.25)
= 100.45

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