SmartphonesBased on an LG smartphone survey, assume that 51% of adults with smartphones use them in theatres. In a separate survey of 250 adults with smartphones, it is found that 109 use them in theatres.

a.If the 51% rate is correct, find the probability of getting 109 or fewer smartphone owners who use them in theatres.

b.Is the result of 109 significantly low?

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,

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

And

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

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,

$$\begin{gathered} \mu =np \\ =250\times 0.51 \\ =127.5 \end{gathered}$$

The standard deviation is,

$$\begin{gathered} \sigma =\sqrt{npq} \\ =\sqrt{250\times 0.51\times 0.49} \\ =7.9041 \end{gathered}$$

a.

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

Here

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

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

Find z-score using$x=109.5$,$\mu =127.5,\sigma =7.90$as follows:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{109.5-127.5}{7.90} \\ =-2.28 \end{gathered}$$

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 $P\left(X\text{or}\text{fewer}\right)⩽0.05$, then it is significantly low.

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