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vredina [299]
3 years ago
9

A company produces a women's bowling ball that is supposed to weigh exactly 14 pounds. Unfortunately, the company has a problem

with the variability of the weight. In a sample of 11 of the bowling balls the sample standard deviation was found to be 0.71 pounds. Construct a 95% confidence interval for the variance of the bowling ball weight. Assume normality.
Mathematics
1 answer:
blsea [12.9K]3 years ago
7 0

Answer: 0.2461

Step-by-step explanation:

Given : A company produces a women's bowling ball that is supposed to weigh exactly 14 pounds.

Sample size : n=11

Degree of freedom =n-1=10

Sample standard deviation : s= 0.71 pounds

Significance level for 95% confidence interval :\alpha=1-0.95=0.05

We assume that the bowling ball weight is normally distributed.

Using chi-square distribution table, the required critical values are :-

\chi^2_{df, \alpha/2}=20.48

\chi^2_{df, 1-\alpha/2}=3.25

Then, the 95% confidence interval for the variance of the bowling ball weight will be :

\dfrac{s^2(n-1)}{\chi_{\alpha/2}}

=\dfrac{(0.71)^2(10)}{20.48}

∴ The 95% confidence interval for the variance of the bowling ball weight will be : 0.2461

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If the average weight for cats is 12
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Answer:

3 standard deviations

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Step-by-step explanation:

The above question is solved using empirical rule

99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ

From the above question

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σ = 2 pounds

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3 standard deviations from the mean is calculated as

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