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Ulleksa [173]
4 years ago
11

Troista Mobile Accessories sells mobile apps on their Web site. If a customer spends on average, $12 per visit and visits the We

b site 20 times each year, what is the average nondiscounted gross profit during a customer's lifetime? Given that Troista makes a margin of 60 percent on the average bill, with 25 percent of customers not returning each year.
Mathematics
1 answer:
Leno4ka [110]4 years ago
8 0

Answer: The required average is $576.

Step-by-step explanation:

Since we have given that

Percent of customers not returning each year = 25%

Average customer lifetime = \dfrac{1}{0.25}=4\ years

Percent of margin on the average bill = 60%

Cost per visit = $12

Number of times each year visited = 20 times

So, Average expense by each customer per year = 12\times 20=\$240

Average margin = 60% of 240 = 0.6\times 240=\$144

Average non discounted gross profit during a customer's lifetime is given by

144\times 4=\$576

Hence, the required average is $576.

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a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

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Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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}

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The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 18.6, \sigma = 5.9

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

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33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768

This probability is 1 subtracted by the pvalue of Z when X = 20.4. So

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

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Z = \frac{X - \mu}{s}

Z = \frac{20.4 - 18.6}{0.6768}

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Z = 2.66 has a pvalue of 0.9961

1 - 0.9961 = 0.0039

0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

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