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Artemon [7]
1 year ago
5

or the past 50 days, daily sales of laundry detergent in a large grocery store have been recorded (to the nearest 10). Units Sol

d 30, 40, 50, 60, 70 Number of Times 9, 12, 15, 10, 5 a. Determine the relative frequency for each number of units sold b. Suppose that the following random number were obtained .12 .96 .53 .80 .95 .10 .40 .45 .77 .29 Use these random numbers to simulate the 10 days of sales
Mathematics
1 answer:
bagirrra123 [75]1 year ago
3 0

Answer, step-by-step explanation:

A. With the previous exercise we can deduce that there is the situation of a number of sales in a grocery store, the relative frequency for the number of units sold, is shown below:

units sold. relative frequency. Acumulative frequency. interval of random numbers

30. 0.16. 0.16. 0.00 <0.16

40. 0.24. 0.4. 0.16 <0.4

50. 0.3. 0.7. 0.4 <0.7

60. 0.2. 0.9. 0.7<09

70. 0.1. 1. 0.9<1

B. For the next point, they give us some random numbers and then it is compared with the simulation of 10 days in sales:

random Units

number. sold

0.12. 30

0.96. 70

0.53. 50

0.80. 60

0.95. 70

0.10. 30

0.40. 50

0.45. 50

0.77. 60

0.29. 40

the two lists are compared so that opposite each one is the result of the simulation

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At Central High School, there are 48 football players and 20 cheerleaders. What is the ratio of cheerleaders to football players
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Answer:

5 : 12

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Cheerleaders : football players

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Suppose that 20% of the adult women in the United States dye or highlight their hair. We would like to know the probability that
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Answer:

71.08% probability that pˆ takes a value between 0.17 and 0.23.

Step-by-step explanation:

We use the binomial approxiation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.2, n = 200. So

\mu = E(X) = np = 200*0.2 = 40

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.2*0.8} = 5.66

In other words, find probability that pˆ takes a value between 0.17 and 0.23.

This probability is the pvalue of Z when X = 200*0.23 = 46 subtracted by the pvalue of Z when X = 200*0.17 = 34. So

X = 46

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

Z = \frac{46 - 40}{5.66}

Z = 1.06

Z = 1.06 has a pvalue of 0.8554

X = 34

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

Z = \frac{34 - 40}{5.66}

Z = -1.06

Z = -1.06 has a pvalue of 0.1446

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Answer:

$1340.49

Step-by-step explanation:

From the attached image

Subtotal Cost of Purchased Items

= (20  \times 16.69)+(2 \times 20.78)+(2  \times 15.58)+(6  \times 21.38)\\+(118  \times 6.60)+(500  \times 0.22)+(10  \times 8.44)+(1  \times 150)\\\\=\$1658

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Tax=$63.83

Therefore, the total bill:

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