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Vladimir [108]
3 years ago
10

This is used for the next few questions: The rating for the new scary movie has a scale of 0 to 10. The average response was tha

t the regular movie attendant enjoyed the movie with 8.3 points and a standard deviation of 0.5 points. What is the percent of people who gave the movie a rating between 6.8 and 8.8? (Write the number as a percent only without a percent sign.)
Mathematics
1 answer:
Pie3 years ago
4 0

Answer:

The percentage that of people who gave the movie a rating between 6.8 and 8.8

<em>P(6.8≤X≤8.8)  = 83.9≅ 84 percentage</em>

Step-by-step explanation:

<u> Step(i):-</u>

Mean of the Population = 8.3 points

Standard deviation of the Population = 0.5 points

Let 'X' be the random variable in normal distribution

<em>Let X = 6.8</em>

Z = \frac{x-mean}{S.D} = \frac{6.8-8.3}{0.5} = -3

Let X = 8.8

Z = \frac{x-mean}{S.D} = \frac{8.8-8.3}{0.5} = 1

The probability that of people who gave the movie a rating between 6.8 and 8.8

<em>P(6.8≤X≤8.8) = P(-3≤Z≤1)</em>

                     = P(Z≤1)- P(Z≤-3)

                    =  0.5 + A(1) - ( 0.5 -A(-3))

                   = A(1) + A(3)       (∵A(-3)=A(3)

                  =  0.3413 +0.4986    (∵ From Normal table)

                 = 0.8399

<u><em>Conclusion:-</em></u>

The percentage that of people who gave the movie a rating between 6.8 and 8.8

<em>P(6.8≤X≤8.8)  = 83.9≅ 84 percentage</em>

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tresset_1 [31]

Answer:

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z= \frac{39-34}{2.5}= 2

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

b) P(X

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So one deviation below the mean we have: (100-68)/2 = 16%

c) z= \frac{29-34}{2.5}= -2

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For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

Step-by-step explanation:

For this case we have a random variable with the following parameters:

X \sim N(\mu = 34, \sigma=2.5)

From the empirical rule we know that within one deviation from the mean we have 68% of the values, within two deviations we have 95% and within 3 deviations we have 99.7% of the data.

We want to find the following probability:

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And replacing we got

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For the second case:

P(X

z= \frac{31.5-34}{2.5}= -1

So one deviation below the mean we have: (100-68)/2 = 16%

For the third case:

P(29 < X

And replacing we got:

z= \frac{29-34}{2.5}= -2

z= \frac{36.5-34}{2.5}= 1

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

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