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given,

The conditions for normal distribution:-

• The shape of the distribution must be bell-shaped (symmetric).
• Probability plot: All points must be closer to a straight line

Now, Construct the normal probability plot by using MINITAB.
MINITAB procedure:
Step 1: Choose Graph $>$ Probability Plot.
Step 2: Choose Singles, and then click OK.
Step 3: In Graph variables, enter the column of Period.
Step 4: Click OK.

Observation:

According to the probability plot of the period, the observations are closer to a straight line.

Therefore, according to the given graph, the distribution of the period is approximately normally distributed.

Step 1:
State the null and alternative hypotheses:
The null hypothesis is:

$H_0:渭=23\text{days}$

It is, that the mean lactation period of grey seals does not differ from 23 days.
An alternative hypothesis is :

$H_a:渭鈮�23\text{days}$

It is, the mean lactation period of grey seals differs from 23 days.

Step 2: Decide on a significance level
Here, the significance level is, $伪=0.05$.

Step 3: Now, Compute the value of the test statistic by using the MINITAB.

MINITAB procedure:
Step 1: Choose State $>$Basic Statistics$>$ 1-Sample Z.
Step 2: In Samples, in Column, enter the column of Period.
Step 3: In Standard deviation, enter$3$.
Step 4: Perform hypothesis test, enter the test mear as
Step 5: Check Options, enter the confidence level as
Step 6: Choose not equal in alternative.
Step 7: Click OK in all dialogue boxes.

Step 4:

The critical value is:

From "Table ll Areas under the standard normal curve" the required value of $Z_{\frac{a}{2}}$with $95\%$ confidence level is $卤1.96$.

Now, Find P.value:

From the MINITAB output, the P.value is $0$.

Step 5:

The critical value approach:

The rejection rule is:

If the test statistic value falls in the rejection region, reject the null hypothesis $H_0$.

If the test statistic value does not fall in the rejection region, then does not reject the null hypothesis $H_0$

Here, the test statistic value falls in the rejection region.

Thus, the null hypothesis is rejected at the $5\%$level.

Hence, it is concluded that the test results are statistically significant at a $5\%$level of significance.

If$P鈮�伪$,

and reject the null hypothesis.

Here, the P-value is $0$, which is less than the significance level.

That is,

$P(=0)鈮�伪(=0.05)$.

Thus, the null hypothesis is rejected at the $5\%$ level.

Hence, it is concluded that the test results are statistically significant at a $5\%$ level of significance.

The data provided is sufficient evidence to conclude that the mean location period of grey seals differs from $23daysata5\%$level.