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olga_2 [115]
3 years ago
8

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.)
Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
7 0

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

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