Question

Based on historical data, your manager believes that 40% of the company's orders come from first-time customers. A random sample of 91 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.26 and 0.43? Answer = (Enter your answer as a number accurate to 4 decimal places.)

Answer 1

Answer:

$P(0.26 \leq p \leq 0.43)=0.7204-0.0032=0.7172$

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The population proportion have the following distribution

$p \sim N(p=0.4,\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.4(1-0.4)}{91}}=0.0514)$

And we can solve the problem using the z score on this case given by:

$z=\frac{p_o -p}{\sqrt{\frac{p(1-p)}{n}}}$

We are interested on this probability:

$P(0.26 \leq p \leq 0.43)$

And we can use the z score formula, and we got this:

$P(\frac{0.26 -0.4}{\sqrt{\frac{0.4(1-0.4)}{91}}} \leq Z \leq \frac{0.43 -0.4}{\sqrt{\frac{0.4(1-0.4)}{91}}})$

$P(-2.726 \leq Z \leq 0.584)$

And we can find this probability like this:

$P(-2.726 \leq Z \leq 0.584)=P(Z<0.584)-P(Z<-2.726)=0.7204-0.0032=0.7172$