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In the morning, the population mean blood glucose level is $\mathrm{μ}=4.60\mathrm{mmol}/\mathrm{l}$, and the population S.D of blood glucose levels is $\mathrm{σ}=0.16\mathrm{mmol}/\mathrm{l}$

The formula used: Standard deviation${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$

Here is the sample size $n=60$

As a result, because the sample size of $60$is more than $30$, we can consider the sample to be substantial.

By the application of CLT,

The sample mean $\stackrel{¯}{x}$follows roughly. Given a normal distribution.

Mean ${\mathrm{μ}}_x=\mathrm{μ}=4.60$and Standard deviation ${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$

$$\begin{gathered} =\frac{0.16}{\sqrt{60}} \\ =0.021 \end{gathered}$$

As a result, the sampling means distribution is normal, with a mean of $4.60\mathrm{mmol}/\mathrm{l}$and S.D. $0.021\mathrm{mmol}/\mathrm{l}$

The sample size is shown here. $n=120$

As a result, we can consider the sample to be a large sample because the sample size of $120$is significantly above $30$

By the application of CLT,

Sample average With $\stackrel{¯}{x}$, the distribution is roughly Normal.

Mean and standard deviation are ${\mathrm{μ}}_x=\mathrm{μ}=4.60$and ${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$respectively.

$$\begin{gathered} =\frac{0.16}{\sqrt{120}} \\ =0.015 \end{gathered}$$

As a result, the sampling mean distribution is normal, with a mean of $4.60\mathrm{mmol}/\mathrm{l}$ and S.D.$0.015\mathrm{mmol}/\mathrm{l}$

No, for samples of size $60$and $120$, it is not necessary to assume that blood glucose levels are distributed regularly. Because, regardless of the population distribution, sample means are normally distributed with large samples (i.e., sample size $>30$), sample means are normally distributed with large samples mean ${\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}$and S.D. ${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$where $\mathrm{μ}=Populationmean,\mathrm{σ}=PopulationS.Dandn=Samplesize$