Question

A consumer mobility report indicates that in a typical day, 51% of users of mobile phones use their phone at least once per hour, 25% use their phone a few times per day, 8% use their phone morning and evening, and 12% hardly ever use their phones. The remaining 4% indicated that they did not know how often they used their mobile phone. Consider a sample of 150 mobile phone users. (a) What is the probability that at least 70 use their phone at least once per hour? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.)

Answer 1

0.8888 = 88.88% probability that at least 70 use their phone at least once per hour

Explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that
`\mu = E(X)`,
`\sigma = √(V(X))`.

51% of users of mobile phones use their phone at least once per hour.

This means that
`p = 0.51`

Consider a sample of 150 mobile phone users.

This means that
`n = 150`

So

`\mu = E(X) = np = 150*0.51 = 76.5`

`\sigma = √(V(X)) = √(np(1-p)) = √(150*0.51*0.49) = 6.12`

What is the probability that at least 70 use their phone at least once per hour?

This is 1 subtracted by the pvalue of Z when X = 70-1 = 69(at least 70 is more than 69). So

`Z = (X - \mu)/(\sigma)`

`Z = (69 - 76.5)/(6.12)`

`Z = -1.22`

`Z = -1.22` has a pvalue of 0.1112

1 - 0.1112 = 0.8888

0.8888 = 88.88% probability that at least 70 use their phone at least once per hour