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?

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

Step-by-step 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:

The standard deviation of the binomial distribution is:

Normal probability distribution

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

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

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 , .

In this problem, we have that:

So

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

Using continuity correction, this is , which is the pvalue of Z when X = 30.5 subtracted by the pvalue of Z when X = 19.5.

X = 30.5

has a pvalue of 0.7123.

X = 19.5

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

here is the

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