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The monthly utility bills in a city are normally distributed, with a mean of $100 and a standard deviation of $12. Find

the probability that a randomly selected utility bill is (a) less than $70, (b) between $89 and $110, and (c) more than
$140.
(a) The probability that a randomly selected utility bill is less than $70 is
(Round to four decimal places as needed.)
(b) The probability that a randomly selected utility will is between $89 and $110 is
(Round to four decimal places as needed.)
(c) The probability that a randomly selected utility bill is more than $140 is
(Round to four decimal places as needed.)

User Hammerbot
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1 Answer

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Final answer:

The probability that a randomly selected bill is less than $70 is 0.0062; between $89 and $110 is 0.1184; and more than $140 is approximately 0.0004. These probabilities are found using Z-scores and the standard normal distribution table.

Step-by-step explanation:

To solve these problems, we use the Z-score formula which helps us find the probability for a normally distributed variable. The Z-score formula is Z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation.

(a)

To find the probability that a randomly selected utility bill is less than $70, we calculate the Z-score for $70:

Z = ($70 - $100) / $12 = -2.5

We then look up this Z-score in the standard normal distribution table to find the probability. The probability associated with Z = -2.5 is approximately 0.0062, which to four decimal places is 0.0062.

(b)

To find the probability that a randomly selected utility bill is between $89 and $110, we calculate the Z-scores for both values:

Z for $89 = ($89 - $100) / $12 = -0.9167

Z for $110 = ($110 - $100) / $12 = 0.8333

The probabilities associated with these Z-scores are for Z = -0.9167 is approximately 0.1793 and for Z = 0.8333 is approximately 0.2977. We subtract the smaller area from the larger to get the probability of being between these two values.

Probability = 0.2977 - 0.1793 = 0.1184

(c)

To determine the probability that a randomly selected utility bill is more than $140, we find the Z-score for $140:

Z = ($140 - $100) / $12 = 3.3333

The probability associated with a Z-score of 3.3333 is very small, near zero. Thus, the probability of a bill over $140 is approximately 0.0004 to four decimal places.

User Pouria Almassi
by
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