Step 1

Given:

This mean recorded temperature is: \(\displaystyle\overline{{x}}={85.69}{F}^{\circ}\)

The population's standard deviation is: \(\displaystyle\sigma={13.60}{F}^{\circ}\)

The number of random samples is: \(\displaystyle{n}={66}\) dates.

Confidence level \(\displaystyle-{90}\%\):

The expression for the confidence level is,

\(\displaystyle{1}-{2}\alpha={0.9}\)

\(\displaystyle\alpha=\frac{{{1}-{0.9}}}{{2}}\)

\(\displaystyle={0.05}\)

Step 2

From the graph of the normal distribution, the confidence level is symmetric about the P axis. The expression to calculate the corresponding value of P is,

\(\displaystyle{P}={1}-\alpha\)

\(\displaystyle={1}-{0.05}\)

\(\displaystyle={0.95}\)

From the z-table corresponding to \(\displaystyle{P}={0.95}\), the value of z is 1.65

The expression to calculate the interval is,

\(\displaystyle{c}.{I}=\overline{{x}}\pm{z}\frac{\sigma}{\sqrt{{n}}}\)

Substitute the values in the above expression.

\(\displaystyle{c}.{l}={85.69}\pm{1.65}\times\frac{13.60}{\sqrt{{66}}}\)

\(\displaystyle={85.69}\pm{2.762}\)

\(\displaystyle={\left({85.69}-{2.762},{85.69}-{2.762}\right)}\)

\(\displaystyle={\left({82.928},{88.452}\right)}\)

Therefore, the interval for confidence leven \(90\%\) is from 82.928 to 88.452

Step 3

Confidence level \(- 95\%:\)

The expression for the confidence level is,

\(\displaystyle{1}-{2}\alpha={0.95}\)

\(\displaystyle\alpha=\frac{{{1}-{0.95}}}{{2}}\)

\(\displaystyle={0.025}\)

From the graph of the normal distribution, the confidence level is symmetric about the P axis. The expression to calculate the corresponding value of P is,

\(\displaystyle{P}={1}-\alpha\)

\(\displaystyle={1}-{0.025}\)

\(\displaystyle={0.975}\)

From the z-table corresponding to \(\displaystyle{P}={0.975}\), the value of z is 1.96.

The expression to calculate the interval is,

\(\displaystyle{c}.{l}=\overline{{x}}\pm{z}\frac{\sigma}{\sqrt{{n}}}\)

Substitute the values in the above expression.

\(\displaystyle{c}.{l}-{85.69}\pm{1.96}\times\frac{13.60}{\sqrt{{66}}}\)

\(\displaystyle={85.69}\pm{3.281}\)

\(\displaystyle={\left({85.69}-{3.281},{85.69}-{3.281}\right)}\)

\(\displaystyle={\left({82.409},{88.971}\right)}\)

Hence, the interval for confidence level \(95\%\) is from 82.409 to 88.971, and the interval for confidence level \(95\%\) is the widest one.

Given:

This mean recorded temperature is: \(\displaystyle\overline{{x}}={85.69}{F}^{\circ}\)

The population's standard deviation is: \(\displaystyle\sigma={13.60}{F}^{\circ}\)

The number of random samples is: \(\displaystyle{n}={66}\) dates.

Confidence level \(\displaystyle-{90}\%\):

The expression for the confidence level is,

\(\displaystyle{1}-{2}\alpha={0.9}\)

\(\displaystyle\alpha=\frac{{{1}-{0.9}}}{{2}}\)

\(\displaystyle={0.05}\)

Step 2

From the graph of the normal distribution, the confidence level is symmetric about the P axis. The expression to calculate the corresponding value of P is,

\(\displaystyle{P}={1}-\alpha\)

\(\displaystyle={1}-{0.05}\)

\(\displaystyle={0.95}\)

From the z-table corresponding to \(\displaystyle{P}={0.95}\), the value of z is 1.65

The expression to calculate the interval is,

\(\displaystyle{c}.{I}=\overline{{x}}\pm{z}\frac{\sigma}{\sqrt{{n}}}\)

Substitute the values in the above expression.

\(\displaystyle{c}.{l}={85.69}\pm{1.65}\times\frac{13.60}{\sqrt{{66}}}\)

\(\displaystyle={85.69}\pm{2.762}\)

\(\displaystyle={\left({85.69}-{2.762},{85.69}-{2.762}\right)}\)

\(\displaystyle={\left({82.928},{88.452}\right)}\)

Therefore, the interval for confidence leven \(90\%\) is from 82.928 to 88.452

Step 3

Confidence level \(- 95\%:\)

The expression for the confidence level is,

\(\displaystyle{1}-{2}\alpha={0.95}\)

\(\displaystyle\alpha=\frac{{{1}-{0.95}}}{{2}}\)

\(\displaystyle={0.025}\)

From the graph of the normal distribution, the confidence level is symmetric about the P axis. The expression to calculate the corresponding value of P is,

\(\displaystyle{P}={1}-\alpha\)

\(\displaystyle={1}-{0.025}\)

\(\displaystyle={0.975}\)

From the z-table corresponding to \(\displaystyle{P}={0.975}\), the value of z is 1.96.

The expression to calculate the interval is,

\(\displaystyle{c}.{l}=\overline{{x}}\pm{z}\frac{\sigma}{\sqrt{{n}}}\)

Substitute the values in the above expression.

\(\displaystyle{c}.{l}-{85.69}\pm{1.96}\times\frac{13.60}{\sqrt{{66}}}\)

\(\displaystyle={85.69}\pm{3.281}\)

\(\displaystyle={\left({85.69}-{3.281},{85.69}-{3.281}\right)}\)

\(\displaystyle={\left({82.409},{88.971}\right)}\)

Hence, the interval for confidence level \(95\%\) is from 82.409 to 88.971, and the interval for confidence level \(95\%\) is the widest one.