Mathematics, 20.03.2020 09:54 danielle4124

A normal distribution is observed from the number of customers retained per company's locations for a certain pharmaceutial company. If the mean is 16 customers and the standard deviation is 2 customers, what is the probability that a randomly selected location, the company retained between 12 and 20 customers

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Answer from: songulakabulut1992

P = 0.9544

Step-by-step explanation:

To find the probability that a randomly selected location, the company retained between 12 and 20 customers, we are going to standardize 12 and 20 using the following equation:

$z=\frac{x-m}{s}$

Where x is the value, m is the mean and s is the standard deviation, so in a standard normal distribution, 12 and 20 are equal to:

$\frac{12-16}{2}=-2\\\\\frac{20-16}{2} =2$

Now, the probability that the company retained between 12 and 20 customers is equal to the probability that the z value is between -2 and 2 in a standard normal distribution. So, using the normal distribution table we have:

P(-2<z<2) = P(z<2) - P(z<-2) = 0.9772 - 0.0228 = 0.9544

Answer from: kellysmith45

Probability that a randomly selected location, the company retained between 12 and 20 customers is 0.9545.

Step-by-step explanation:

We are given that a normal distribution is observed from the number of customers retained per company's locations for a certain pharmaceutical company where the mean is 16 customers and the standard deviation is 2 customers.

Let X = number of customers retained per company's locations for a certain pharmaceutical company

So, X ~ N( $\mu=16,\sigma^{2}=2^{2}$ )

Now, the z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean

$\sigma$ = standard deviation

So, probability that a randomly selected location, the company retained between 12 and 20 customers is given by = P(12 < X < 20)

P(12 < X < 20) = P(X < 20) - P(X $\leq$ 12)

P(X < 20) = P( $\frac{X-\mu}{\sigma}$ < $\frac{20-16}{2}$ ) = P(Z < 2) = 0.97725

P(X $\leq$ 12) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{12-16}{2}$ ) = P(Z $\leq$ -2) = 1 - P(Z < 2)

= 1 - 0.97725 = 0.02275

Now, P(12 < X < 20) = 0.97725 - 0.02275 = 0.9545

Therefore, probability that a randomly selected location, the company retained between 12 and 20 customers is 0.9545.

Answer from: Quest

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Answer from: Quest

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