Question

A recent study reported that 28% of shoppers only review one page when searching online for product information. A random sample of 100 shoppers was randomly selected. What is the probability that between 20 and 30 of these shoppers only review one page when searching online?

Answer 1

68.29% probability that between 20 and 30 of these shoppers only review one page when searching online

Explanation:

I am going to use the normal approximation to the binomial to solve this question.

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))`.

In this problem, we have that:

`n = 100, p = 0.28`

So

`\mu = E(X) = np = 100*0.28 = 28`

`\sigma = √(V(X)) = √(np(1-p)) = √(100*0.28*0.72) = 4.49`

What is the probability that between 20 and 30 of these shoppers only review one page when searching online?

Using continuity correction, this is
`P(20 - 0.5 \leq X \leq 30 + 0.5) = P(19.5 \leq X \leq 30.5)`, which is the pvalue of Z when X = 30.5 subtracted by the pvalue of Z when X = 19.5.

X = 30.5

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

`Z = (30.5 - 28)/(4.49)`

`Z = 0.56`

`Z = 0.56` has a pvalue of 0.7123.

X = 19.5

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

`Z = (19.5 - 28)/(4.49)`

`Z = -1.89`

`Z = -1.89` has a pvalue of 0.0294

0.7123 - 0.0294 = 0.6829

68.29% probability that between 20 and 30 of these shoppers only review one page when searching online