91Ó°ÊÓ

The amount that households pay for Internet access varies significantly, but the average monthly rate is $50 and the standard deviation is $20.

Although you are aware that the population distribution is not normal, you are unaware of the actual population distribution.

It is consequently impossible to compute the probability for a single randomly picked household because doing so necessitates knowing the exact population distribution.

Given,

Formula used:

${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$

Let calculate

The amount that households pay for Internet access varies significantly, but the average monthly rate is $50 and the standard deviation is $20.

The centre limit theorem states that if the sample size is bigger than 30, the sample mean$\stackrel{¯}{x}$sampling distribution is approximately normal.

The central limit theorem can be used, and hence the sampling distribution of the sample mean is approximately normal.

Given,

$$\begin{gathered} \mathrm{μ}=50 \\ \mathrm{σ}=20 \\ n=50 \end{gathered}$$

The value of z-score is

$z=\frac{\stackrel{¯}{x}-{\mathrm{μ}}_{\stackrel{¯}{x}}}{{\mathrm{σ}}_{\stackrel{¯}{x}}}=\frac{\stackrel{¯}{x}-\mathrm{μ}}{\mathrm{σ}/\sqrt{n}}=\frac{55-50}{20/\sqrt{50}}=1.77$

The standard probability is

$$\begin{gathered} P(\stackrel{¯}{X}â‰\yen 55)=P(Z>1.77) \\ =1-P(z<1.77) \\ =1-0.9616 \\ =0.0384 \\ =3.84\% \end{gathered}$$