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.