Question

According to the U.S. Energy Administration, the mean monthly household electric bill in the United States is $99.70 with a standard deviation of $20. Given that household electric bills follow a Normal Distribution, what is the probability that a randomly chosen household electric bill:
A) is over $131.50? B) is less than $80? C) is between $85.50 and $111.50? D) is either higher than $105 or less than $90?

Answer 1

Answer:

A. 5.82%

B. 16.11%

C. 54.93%

D. 29.42%

Step-by-step explanation:

First draw the normal distribution curve for the situation as in the attached image. The middle line is the mean(μ) of 99.7, and every line away from the middle is a standard deviation(σ) of 20 more or less. Next calculate the z-score for each scenario. The z-score basically calculates how many standard deviations, the number in question is, from the mean.

Z-Scores:

The formula to calculate a z-score is as follows:

$z = \frac{x-mean}{standard deviation}$

A:  $\frac{131.5-99.7}{20} =1.57$

B:  $\frac{80-99.7}{20} =-0.99$

C:  $\frac{85.5-99.7}{20} =-0.71$

   $\frac{111.5-99.7}{20} =0.79$

D:  $\frac{90-99.7}{20} =-0.49$

  $\frac{105-99.7}{20} =0.27$

Convert Z scores to percentiles:

Use the attached tables to convert z scores to percentiles. The tables are in decimals, so multiply by 100 to find percentile, and note that percentages are given to the left of the vertical line.

A: 94.18 % to the left of the line, which is less than 131.50, thus 5.82% is more.

B: 16.11 % is less than 80.

C: 23.89 % is less than 85.50

    78.52% is less than 115.50

    Between 85.50 and 115.50 can be calculated as $78.52-23.59=54.93$

D: 31.21 % is less than 90

    60.64% is less than 105

    Between 85.50 and 115.50 can be calculated as $60.64-31.21=29.43$