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Lubov Fominskaja [6]
3 years ago
14

Electricity bills: According to a government energy agency, the mean monthly household electricity bill in the United States in

2011 was $109.55. Assume the amounts are normally distributed with standard deviation $21.00. Use the TI-84 Plus calculator to answer the following. (a) What proportion of bills are greater than $135? (b) What proportion of bills are between $84 and $143? (c) What is the probability that a randomly selected household had a monthly bill less than $119? Round the answers to at least four decimal places.
Mathematics
1 answer:
xxMikexx [17]3 years ago
8 0

Answer:

a) 0.11277

b) 0.83253

c) 0.67364

Step-by-step explanation:

The formula for calculating a z-score is is z = (x-μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation.

From the question,

Mean = $109.55.

Standard deviation = $21.00.

a) What proportion of bills are greater than $135?

z = (x-μ)/σ

z = 135 - 109.55/21

= 1.2119

Probability value from Z-Table:

P(x<135) = 0.88723

P(x>135) = 1 - P(x<135) = 0.11277

The proportion of bills are greater than $135 = 0.11277

(b) What proportion of bills are between $84 and $143?

For x = $84

z = (x-μ)/σ

z = 84 - 109.55/21

= -1.21667

Probability value from Z-Table:

P(x = 84) = 0.11187

For x = 143

z = (x-μ)/σ

z = 143 - 109.55/21

= 1.59286

Probability value from Z-Table:

P(x = 143) = 0.9444

The proportion of bills are between $84 and $143

= P(x = $143) - P(x = 84)

= 0.9444 - 0.11187

= 0.83253

(c) What is the probability that a randomly selected household had a monthly bill less than $119?

For x = $119

=119 - 109.55/ 21

= 0.45

Probability value from Z-Table:

P(x<119) = 0.67364

The probability that a randomly selected household had a monthly bill less than $119 is 0.67364

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