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Sidana [21]
3 years ago
6

Suppose Kristen is researching failures in the restaurant business. In the city where she lives, the probability that an indepen

dent restaurant will fail in the first year is 32 % . She obtains a random sample of 72 independent restaurants that opened in her city more than one year ago and determines if each one had closed within a year. What are the mean and standard deviation of the number of restaurants that failed within a year? Please give your answers precise to two decimal places.
Mathematics
1 answer:
WINSTONCH [101]3 years ago
4 0

Answer:

The mean of the number of restaurants that failed within a year is 23.04 and the standard deviation is 3.96.

Step-by-step explanation:

For each restaurant, there are only two possible outcomes. Either it fails during the first year, or it does not. The probability of a restaurant failling during the first year is independent of other restaurants. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

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

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)}

The probability that an independent restaurant will fail in the first year is 32%.

This means that p = 0.32

72 independent restaurants

This means that n = 72

Mean:

E(X) = np = 72*0.32 = 23.04

Standard deviation:

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{72*0.32*0.68} = 3.96

The mean of the number of restaurants that failed within a year is 23.04 and the standard deviation is 3.96.

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Hope this helps.

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6 0
3 years ago
The data show the average monthly temperatures for two cities over a 6-month period. City 1: {20, 24, 40, 63, 76, 89} City 2: {4
Mrac [35]

Answer:

The MAD of city 2 is <u>less than</u> the MAD for city 1, which means the average monthly temperature of city 2 vary <u>less than</u> the average monthly temperatures for City 1.

Step-by-step explanation:

For comparing the mean absolute deviations of both data sets we have to calculate the mean absolute deviation for both data sets first,

So for city 1:

Mean = x1 = \frac{20+24+40+63+76+89}{6}

x1 = \frac{312}{6}

x1 = 52

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored as the deviation is the distance of value from the mean and it cannot be negative. For this purpose absolute is used)

20-52 = -32=32\\24-52=-28=28\\40-52=-12=12\\63-52=11\\76-52=24\\89-52=37

The deviations will be added then.Mean Absolute Deviation = \frac{32+28+12+11+24+37}{6} \\=\frac{144}{6}\\=24

So the mean absolute deviation for city 1 is 24 ..

For city 2:

Mean = x2 = \frac{41+50+58+62+72+83}{6}

x2 = \frac{366}{6}

x2 = 61

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored)

41-61=-20=20\\50-61=-11=11\\58-61=-3=3\\62-61=1\\72-61=11\\83-61=22

The deviations will be added then.Mean Absolute Deviation = \frac{20+11+3+1+11+22}{6} \\=\frac{68}{6}\\=11.33

So the MAD for city 2 is 11.33 ..

So,

The MAD of city 2 is <u>less than</u> the MAD for city 1, which means the average monthly temperature of city 2 vary <u>less than</u> the average monthly temperatures for City 1.

3 0
3 years ago
Read 2 more answers
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