Mathematics, 18.03.2020 00:08 isaacbrown7one

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?

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Answer from: alessandrotabares

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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Answer from: christyr2002

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

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

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

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 from: Quest

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Answer from: Quest

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