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JulsSmile [24]
3 years ago
7

MS. Kitts works at a music store. Last week she sold 6 more than 3 times the number of CDs that she sold this week. Ms. Kitts so

ld a total of 110 CDs over the 2 weeks. Which system of equations can be used to find x, the number of CDs she sold last week, and y, the number of CDs she sold this week?
Mathematics
1 answer:
puteri [66]3 years ago
6 0

x - 3y = 6 and x + y = 110 represents the system of equations used to find x, the number of CDs she sold last week, and y, the number of CDs she sold this week

number of CDs Kitts sold last week "x" is 84 and number of CDs she sold this week "y" is 26

<h3><u>Solution:</u></h3>

Let "x" be the number of CDs Kitts sold last week

Let "y" be the number of CDs she sold this week

<em><u>Given that Last week she sold 6 more than 3 times the number of CDs that she sold this week</u></em>

Number of CDs Kitts sold last week = 6 + 3(number of CDs that she sold this week)

x = 6 + 3y ----- eqn 1

x - 3y = 6 ------- eqn 2

<em><u>Given that Ms. Kitts sold a total of 110 CDs over the 2 weeks</u></em>

Number of CDs Kitts sold last week + number of CDs she sold this week = 110

x + y = 110 ----- eqn 3

Thus eqn 2 and eqn 3 represents the system of equations used to find x, the number of CDs she sold last week, and y, the number of CDs she sold this week

<em><u>Let us solve the above equations to find values of "x" and "y"</u></em>

Substitute eqn 1 in eqn 3

6 + 3y + y = 110

6 + 4y = 110

4y = 104

<h3>y = 26</h3>

From eqn 3,

x + 26 = 110

x = 110 - 26 = 84

<h3>x = 84</h3>

Thus number of CDs Kitts sold last week = x = 84

Number of CDs she sold this week = y = 26

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a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b) 0.25% probability that his average kicks is less than 36 yards

c) 0.11% probability that his average kicks is more than 41 yards

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d-b) 1.32% probability that his average kicks is less than 36 yards

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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}

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.

Central Limit Theorem

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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 38.4, \sigma = 5.4, n = 40, s = \frac{5.4}{\sqrt{40}} = 0.8538

a. What is the distribution of the sample mean? Why?

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b. What is the probability that his average kicks is less than 36 yards?

This is the pvalue of Z when X = 36. So

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

By the Central Limit Theorem

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

Z = \frac{36 - 38.4}{0.8538}

Z = -2.81

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c. What is the probability that his average kicks is more than 41 yards?

This is 1 subtracted by the pvalue of Z when X = 41. So

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

Z = \frac{41 - 38.4}{0.8538}

Z = 3.05

Z = 3.05 has a pvalue of 0.9989

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0.11% probability that his average kicks is more than 41 yards

d. If the sample size is 25 in the above problem, what will be your answer to part (a) , (b)and (c)?

Now n = 25, s = \frac{5.4}{\sqrt{25}} = 1.08

So

a)

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

b)

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

Z = \frac{36 - 38.4}{1.08}

Z = -2.22

Z = -2.22 has a pvalue of 0.0132

1.32% probability that his average kicks is less than 36 yards

c)

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

Z = \frac{41 - 38.4}{1.08}

Z = 2.41

Z = 2.41 has a pvalue of 0.9920

1 - 0.9920 = 0.0080

0.80% probability that his average kicks is more than 41 yards

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