91Ó°ÊÓ

The number of adults with smartphones is recorded. The given sample size$n=250$ and probability of success $p=0.51$.

Then,

$$\begin{gathered} q=1-p \\ =1-0.51 \\ =0.49 \end{gathered}$$

Let X be the number of adults who use smartphones in theatres.

From the given information,

$$\begin{gathered} np=250×0.51 \\ =127.5 \\ >5 \end{gathered}$$

And

$$\begin{gathered} nq=250×0.49 \\ =122.5 \\ >5 \end{gathered}$$

Here both and are greater than 5. Hence probabilities from a binomial

probability distribution can be approximated reasonably well by using a normal distribution.

The mean value is,

$$\begin{gathered} \mathrm{μ}=np \\ =250×0.51 \\ =127.5 \end{gathered}$$

The standard deviation is,

$$\begin{gathered} \mathrm{σ}=\sqrt{npq} \\ =\sqrt{250×0.51×0.49} \\ =7.9041 \end{gathered}$$

a.

The probability of 109 or fewer smartphone owners who use their smartphones in theatres is expressed using the continuity correction$P\left(Xâ\copyright ½n\right)=P\left(X<n+0.5\right)$

Here

$$\begin{gathered} P\left(Xâ\copyright ½109\right)=P\left(X<109+0.5\right) \\ =P\left(X<109.5\right) \end{gathered}$$

Thus, the probability is expressed as,$P\left(X<109.5\right)$

Find z-score using$x=109.5$,$\mathrm{μ}=127.5,\mathrm{σ}=7.90$as follows:

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{109.5-127.5}{7.90} \\ =-2.28 \end{gathered}$$

The z-score is -2.28.

Using the standard normal table, the cumulative area to the left of -2.28 is 0.0113.

b.

Significantly low events have probability 0.05 or less for fewer and x events.

If $P\left(X\text{or}\text{fewer}\right)â\copyright ½0.05$, then it is significantly low.

As, 0.0113 < 0.05, therefore it is significantly low.