Question

Results of a survey revealed that the distribution of the amount of the monthly utility bill of a 3-bedroom house using gas or electric energy had a mean of $97 and a standard deviation of $12.

If the distribution of monthly bills can be considered mound-shaped and symmetric, what percentage of homes will have a monthly bill of more than $85 and less than $109?

If nothing is known about the shape of the distribution of monthly bills, what percentage of homes will have a monthly bill between $61 and $133?

If the distribution of monthly bills can be considered mound-shaped and symmetric, what percentage of homes will have a monthly bill of more than $121?

use Excel

explain how to calculate standard deviation and how you got the percentage?

Answer 1

Here the mean monthly bill = $ 97

standard deviaiton of monthly bill = $ 12

now it is given that the distribution of monthly bills can be considered mound-shaped and symmetric, which can be assumed as normal distribution.

so here if x is the monthly bill of a random month. we have to find

Pr($ 85 < x < $ 109) = Pr(x < $ 109 ; $ 97; $ 12) - Pr(x < $ 85 ; $ 97 ; $ 12)

Z₂ = (109 - 97)/12 = 1

Z₁  = (85 - 97)/12 = -1

Pr($ 85 < x < $ 109) = NORMSIDT(1) - NORMSDIST(-1) = 0.8413 - 0.1587 = 0.6827 or 68.27%

Now as we can solve by the empirical rule that 68% of the values will be withing 1 standard deviation so here values $ 109 and $ 85 are exactly one standard deviation away from mean.

so here 68.3% of monthly bill would be in between $ 85 and $ 109.

(2) If nothing is know about the distribution so we will use the chebyshev identity which entails that

Pr(l x - l < k) > 1 - 1/k²

so here we have to find that what number of values lie between $ 61 and $ 133

so here 61 = 97 - 3 * 12 and 133 = 97 + 3 * 12

that means these values are 3 standard deviation apart.

so here

Pr(l x - l < k) = Pr( l x - 97 l < 36) > 1 - 1/3²

Pr(l x - l < k) = Pr( l x - 97 l < 36) > 0.88889

so here atleast 88.89% chance of homes will have a monthly bill between $61 and $133.

Question 3

If the distribution of monthly bills can be considered mound-shaped and symmetric, what percentage of homes will have a monthly bill of more than $121?

Here again we know that the distribution is mound shaped and symmetric which can be considered normal distribution.

Pr(x > $ 121) = 1 - Pr(x < $ 121; $ 97 ; $ 12)

Z = (121 - 97)/12 = 2

Pr(x > $ 121) = 1 - Pr(x < $ 121; $ 97 ; $ 12) = 1 - NORMSDIST(2) = 1 - 0.97725 = 0.02275 = 2.275%

so gere 2.275% of homes will have a monthly bill of more than $ 121.

Question 4