91Ó°ÊÓ

Let the population mean for hospital stay length be $\mathrm{μ}$days.

Given that the population s. d.,$\mathrm{σ}=8.3$ days, the population distribution of hospital stay lengths is unknown.

Formula used:

Here the sample size is $80$i.e., $n=80$

Because the sample size of $80$is larger than $30$, we can consider the sample to be substantial.

Mean of the sample mean $$\begin{gathered} {\mathrm{μ}}_{\stackrel{¯}{X}} \\ =\mathrm{μ}Days \end{gathered}$$

S.D of sample mean ${\mathrm{σ}}_{\stackrel{¯}{X}}$

$$\begin{gathered} =\frac{\mathrm{σ}}{\sqrt{n}} \\ =\frac{8.3}{\sqrt{80}} \\ =0.92796\text{Days} \end{gathered}$$

As a result of using CLT sampling, the distribution of all potential sample means of size $80$ samples is normal, with a mean of $\mathrm{μ}$ and a standard deviation of $0.92796$ days.

The distribution of hospital stay lengths is right-skewed. Partially, this knowledge does not invalidate the result (a). We can use CLT to approximate the distance of the sample mean as a normal distribution because the sample size is relatively large.

We have to find $P(\mathrm{μ}-2≤\stackrel{¯}{X}≤\mathrm{μ}+2)$

Where $\stackrel{¯}{X}~N\left(\mathrm{μ},{\mathrm{σ}}_{\stackrel{¯}{X}}^2\right)$

Where ${\mathrm{σ}}_{\stackrel{¯}{X}}=\frac{\mathrm{σ}}{\sqrt{n}},n=80$

$$\begin{gathered} =0.92796days \\ ∴P(\mathrm{μ}-2≤\stackrel{¯}{X}≤\mathrm{μ}+2) \\ =P\left(\frac{\mathrm{μ}-2-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}≤\frac{\stackrel{¯}{X}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}≤\frac{\mathrm{μ}+2-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}\right) \\ =P\left(\frac{-2}{0.92796}≤z≤\frac{2}{0.92796}\right),z=\frac{\stackrel{¯}{X}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{X}}}~N(0,1) \\ =P(-2.1553≤z≤2.1583) \\ =\mathrm{Φ}(2.1553)-\mathrm{Φ}(-2.1553) \\ =2\mathrm{Φ}(2.1553)-1 \\ =2×.9844308-1 \\ =0.9660896 \end{gathered}$$

As a result, the probability of guessing the population mean by the value of the sample mean of a sample of size $80$ with a sampling error of at most $2$ days is $0.9660896$