Question

A particular type of Tennis racket comes in a midsize version and an oversize version. Sixty percent of all customers at a certain store want the oversize version. Among 100 randomly selected customers, let X denote the number ofcustomers who want this type of racket. Compute the approximate probability that X is at least 70.

Answer 1

2.07% probability that X is at least 70.

Explanation:

I am going to use the binomial approximation to the normal 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 = 0.6`.

So

`\mu = E(X) = np = 100*0.6 = 60`

`\sigma = √(V(X)) = √(np(1-p)) = √(100*0.6*0.4) = 4.9`

Compute the approximate probability that X is at least 70.

This probability is 1 subtracted by the pvalue of Z when X = 70. So

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

`Z = (70 - 60)/(4.9)`

`Z = 2.04`

`Z = 2.04` has a pvalue of 0.9793

1 - 0.9793 = 0.0207

2.07% probability that X is at least 70.