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The sample number of body temperatures is $n=106$.

The mean body temperature is 98.20.

The sample standard deviation is$s=0.62$.

The level of confidence is 95%.

The degrees of freedom is computed as,

$$\begin{gathered} df=n-1 \\ =106-1 \\ =105 \end{gathered}$$

The level of confidence is 95%, which implies that the level of significance is 0.05.

Using the Chi-square table, the critical values at 0.05 level of significance and at 105 degrees of freedom are localid="1648108100252" ${蠂}_L^2=78.5364$and${蠂}_R^2=135.247$.

The 95% confidence interval estimate of the standard deviation of the body

temperatures for the entire population is computed as,

localid="1648108270737" $$\begin{gathered} \sqrt{\frac{\left(n-1\right)s^2}{{蠂}_R^2}}<蟽<\sqrt{\frac{\left(n-1\right)s^2}{{蠂}_L^2}} \\ \sqrt{\frac{\left(106-1\right){0.62}^2}{135.247}}<蟽<\sqrt{\frac{\left(106-1\right){0.62}^2}{78.5364}} \\ 0.55<蟽<0.72 \end{gathered}$$

Therefore, the 95% confidence interval estimate of the standard deviation of the body temperatures for the entire population is $0.55{}^{\text{o}}\text{F}<蟽<0.72{}^{\text{o}}\text{F}$.