91Ó°ÊÓ

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

Then,

\(\begin{aligned}{c}q = 1 - p\\ = 1 - 0.51\\ = 0.49\end{aligned}\)

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

From the given information,

\(\begin{aligned}{c}np = 250 \times 0.51\\ = 127.5\\ > 5\end{aligned}\)

And

\(\begin{aligned}{c}nq = 250 \times 0.49\\ = 122.5\\ > 5\end{aligned}\)

Here both\({\bf{np}}\)and\({\bf{nq}}\)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,

\(\begin{aligned}{c}\mu = np\\ = 250 \times 0.51\\ = 127.5\end{aligned}\)

The standard deviation is,

\(\begin{aligned}{c}\sigma = \sqrt {npq} \\ = \sqrt {250 \times 0.51 \times 0.49} \\ = 7.9041\end{aligned}\)

a.

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

Here

\(\begin{aligned}{c}P\left( {X \le 109} \right) = P\left( {X < 109 + 0.5} \right)\\ = P\left( {X < 109.5} \right)\end{aligned}\)

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:

\(\begin{aligned}{c}z = \frac{{x - \mu }}{\sigma }\\ = \frac{{109.5 - 127.5}}{{7.90}}\\ = - 2.28\end{aligned}\)

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 \(P\left( {X\;{\rm{or}}\;{\rm{fewer}}} \right) \le 0.05\), then it is significantly low.

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