There has been much media coverage of the high cost of medicinal drugs in the United States. One concern is the large variation from pharmacy to pharmacy. To investigate, a consumer advocacy group took a random sample of 100 pharmacies around the country and recorded the price (in dollars per 100 pills) of Prozac. Compute the range, variance, and standard deviation of the prices. Discuss what these statistics tell you.

Answers

Answer 1

Data for the question :

102.05 99.85 112.3 97.15 111.23 105.37 105.64 106.5 102.97 107.82 106.36 111.24 107.28 114.14 106.28 106.96 98.25 111.55 107.75 101.02 101.12 97.7 97.66 100.54 115.77 112.91 111.04 112.15 102.87 101.14 107.13 108.56 109.56 103.57 108.68 104.59 116.74 116.22 100.22 103.97 111.2 109.34 115.78 101.59 107.93 104.23 96.25 103.84 102.47 102.96 99.26 101.42 108.58 107.69 99.88 102.71 111.25 99.4 117.04 106.35 110.44 102.34 107.25 107.63 105.2 109.14 115.54 101.51 108.49 112.32 109.27 97.54 102.46 105.94 109.42 111.05 102.63 106.99 102.03 108.84 118.8 108.64 95.35 105.47 104.45 102.15 111.4 108.27 104.82 108.4 109.05 116.11 103.7 121.2 99.62 102.81 109.56 103.35 113.02 103.79

Answer:

Range = 25.35

Variance = 29.46

Standard deviation = 5.43

The variation in price of Prozac is high

Step-by-step explanation:

The range of the data :

Maximum - Minimum.

121.2 - 95.35 = 25.35

The variance, s :

s² = Σ(X - m²) / n - 1

Mean, m = Σx / n

X = individual data point

m = mean of data

n = sample size

Using a calculator of save time and ensure accuracy :

s² = 29.45522

The standard deviation, s

s = sqrt(variance)

s = sqrt(s²)

s = sqrt(29.45522)

s = 5.42726.

The range, variance and standard deviation, all measure the degree of variation in a dataset. The values of these statistical measure obtain for the price of 1 product across different pharmaceutical stores, suggests thatvthe variation in price is high;

With a range of about 25.35 and standard deviation of 5.43


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A is august and B is October

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Answer:

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Answer:

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Answer:

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Answers

Answer:

Option E, two-proportion z test  should be used to determine whether these data provide sufficient evidence to reject the hypothesis that the proportion of shoppers at the suburban mall who had been to a movie in the past month is the same as the proportion of shoppers in the large downtown shopping area who had been to a movie in the past month

Step-by-step explanation:

The complete question is

In a random sample of 60 shoppers chosen from the shoppers at a large suburban mall, 36 indicated that they had been to a movie in the past

month. In an independent random sample of 50 shoppers chosen from the shoppers in a large downtown shopping area, 31 indicated that

they had been to a movie in the past month. What significance test should be used to determine whether these data provide sufficient

evidence to reject the hypothesis that the proportion of shoppers at the suburban mall who had been to a movie in the past month is the same

as the proportion of shoppers in a large downtown shopping area who had been to a movie in the past month?

A one-proportion z interval B two-proportion z interval

B two-proportion z interval

C two-sample t test D one-proportion z test

D one-proportion z test

E two-proportion z test

Solution

Two proportion z test is used to compare two proportions. In this test the null hypothesis is that the two proportions are equal and the alternate hypothesis is that the proportions are not the same. The random sample of populations serve as two proportions.

Hence, option E is the best choice of answer

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Answers

Answer:

0.36 = 36% probability that the land has oil and the test predicts it

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

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[tex]P(A \cap B)[/tex] is the probability of both A and B happening.

P(A) is the probability of A happening.

45% chance that the land has oil.

This means that [tex]P(A) = 0.45[/tex]

He buys a kit that claims to have an 80% accuracy rate of indicating oil in the soil.

This means that [tex]P(B|A) = 0.8[/tex]

What is the probability that the land has oil and the test predicts it?

This is [tex]P(A \cap B)[/tex]. So

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

[tex]P(B \cap A) = P(B|A)*P(A) = 0.8*0.45 = 0.36[/tex]

0.36 = 36% probability that the land has oil and the test predicts it

Answer:

The probability that the land has oil and the test predicts it is 36%

Step-by-step explanation:

So option  C.  0.36 is correct for plato users

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