Question

Suppose 3% of the people contacted by phone are receptive to a certain sales pitch and buy your product. If your sales staff contacts 2000 people, what is the probability that more than 100 of the people contacted will purchase your product?

Answer 1

Answer:

0% probability that more than 100 of the people contacted will purchase your product

Step-by-step explanation:

To solve this question, i am going to use the binomial approximation to the normal.

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:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

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:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$p = 0.03, n = 2000$

So

$\mu = E(X) = np = 2000*0.03 = 60$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2000*0.03*0.97} = 7.63$

What is the probability that more than 100 of the people contacted will purchase your product?

This is 1 subtracted by the pvalue of Z when X = 100. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{100 - 60}{7.63}$

$Z = 5.24$

$Z = 5.24$ has a pvalue of 1

1 - 1 = 0

0% probability that more than 100 of the people contacted will purchase your product

Answer 2

The probability that more than 100 purchase the product is 0

The given parameters are:

$\mathbf{p = 3\%}$

$\mathbf{n = 2000}$

Calculate the mean using

$\mathbf{\mu = np}$

$\mathbf{\mu = 2000 \times 3\%}$

$\mathbf{\mu = 60}$

Calculate the standard deviation using

$\mathbf{\sigma = \sqrt{\mu(1 - p)}}$

So, we have:

$\mathbf{\sigma = \sqrt{60 \times (1 - 3\%)}}$

$\mathbf{\sigma = 7.63}}$

For x = 100, the z-score is calculated using:

$\mathbf{z = \frac{x - \mu}{\sigma}}$

So, we have:

$\mathbf{z = \frac{100 - 60}{7.63}}$

$\mathbf{z = \frac{40}{7.63}}$

$\mathbf{z = 5.24}$

So, the probability that more than 100 purchase the product is calculated using:

$\mathbf{P(x > 100) = P(z > 5.24)}$

From z-score of probabilities, we have:

$\mathbf{P(x > 100) = 0}$

Hence, the probability is 0

Read more about probabilities at:

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