91Ó°ÊÓ

The probability that the sample proportion$V$estimates the population proportion $0.56$within $Â\pm 0.04$.

The probability that the sample proportion $V$will accurately predict the population within$Â\pm 0.04$is determined as follows:

$P(0.52≤V≤0.60)=P\left(\frac{0.52-\mathrm{μ}}{\mathrm{σ}}≤\frac{x-\mathrm{μ}}{\mathrm{σ}}≤\frac{0.60-\mathrm{μ}}{\mathrm{σ}}\right)$

Here, population mean is $\left(\mathrm{μ}\right)=0.56$, and the population standard is deviation$\left(\mathrm{σ}\right)=0.019$

$$\begin{gathered} =P\left(\frac{0.52-0.56}{0.019}≤\frac{x-\mathrm{μ}}{\mathrm{σ}}≤\frac{0.60-\mathrm{μ}}{\mathrm{σ}}\right) \\ =P(−2.11≤Z≤2.11) \\ =0.9652 \end{gathered}$$

Let, $72\%$of the respondents said they had voted $(V=0.72).$

To find the probability for $P(Vâ‰\yen 0.72)$to check, could be claimed by the people who did not vote.

By using the normal probability table to calculate the probability as:

$$\begin{gathered} P(Vâ‰\yen 0.72)=P(\frac{x-\mathrm{μ}}{\mathrm{σ}}â‰\yen \frac{0.72-\mathrm{μ}}{\mathrm{σ}}) \\ =P\left(Zâ‰\yen \frac{0.72−0.56}{0.019}\right) \\ =P(zâ‰\yen 8.42) \\ =0.0001 \end{gathered}$$

As a result, the chance is too low. It's possible to make a case for it. Because getting a sample proportion of $72$percent by random is virtually hard if the genuine percentage is $56$percent .