Financial Risk Manager Part 1

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An auto insurance company intends to establish the mean claim amount demanded by policyholders who own SUVs. After extensive analysis of its records, the company believes the standard deviation of such claims is about $200. The company wishes to construct a 95% confidence interval for the mean claim amount such that the interval is of width "±$50". Determine the value of n, the sample size that would be required to achieve this.

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Explanation:

Explanation

For a 95% confidence interval for the mean μ with normal distribution, the formula is:

$\bar{X} \pm 1.96 \frac{\sigma}{\sqrt{n}}$

Given:

Standard deviation σ = $200
Margin of error = $50
Confidence level = 95% (z-score = 1.96)

We need to find n such that:

$`$ 1.96` \times \frac{\sigma}{\sqrt{n}} = 50$$

Substitute σ = 200:

$`$ 1.96` \times \frac{200}{\sqrt{n}} = 50$$

$\frac{392}{\sqrt{n}} = 50$

$`$ 392` = 50\sqrt{n}$$

$\sqrt{n} = \frac{392}{50} = 7.84$

$n = (7.84)^2 = 61.4656 \approx 62$

Therefore, a sample size of 62 is required to achieve a 95% confidence interval with a width of ±$50.

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