To answer these questions, we need to use the normal distribution function on the TI-84 Plus calculator.

Part 1 of 3
(a) What proportion of bills are greater than $131?

First, we need to convert $131 to a z-score. The z-score is calculated as (X - μ) / σ, where X is the value we're interested in, μ is the mean, and σ is the standard deviation. So, the z-score for $131 is (131 - 110.1) / 18 = 1.16.

Next, we use the normalcdf function on the TI-84 Plus to find the proportion of bills greater than $131. We input the z-score we just calculated (1.16), and a large number as the upper limit (like 9999), with the mean (0) and standard deviation (1). The result is the proportion of bills greater than $131.

Part 2 of 3
(b) What proportion of bills are between $81 and $142?

We need to convert both $81 and $142 to z-scores. The z-score for $81 is (81 - 110.1) / 18 = -1.62, and for $142 it is (142 - 110.1) / 18 = 1.77.

We use the normalcdf function again, this time with -1.62 as the lower limit and 1.77 as the upper limit. The result is the proportion of bills between $81 and $142.

Part 3 of 3
(c) What is the probability that a randomly selected household had a monthly bill less than $110?

The z-score for $110 is (110 - 110.1) / 18 = -0.006. We use the normalcdf function with a large negative number as the lower limit (like -9999), and -0.006 as the upper limit. The result is the probability that a randomly selected household had a monthly bill less than $110.

Please note that the actual numbers may vary slightly due to rounding.

Question

To answer these questions, we need to use the normal distribution function on the TI-84 Plus calculator.

Part 1 of 3
(a) What proportion of bills are greater than $131?

First, we need to convert $131 to a z-score. The z-score is calculated as (X - μ) / σ, where X is the value we're interested in, μ is the mean, and σ is the standard deviation. So, the z-score for $131 is (131 - 110.1) / 18 = 1.16.

Next, we use the normalcdf function on the TI-84 Plus to find the proportion of bills greater than $131. We input the z-score we just calculated (1.16), and a large number as the upper limit (like 9999), with the mean (0) and standard deviation (1). The result is the proportion of bills greater than $131.

Part 2 of 3
(b) What proportion of bills are between $81 and $142?

We need to convert both $81 and $142 to z-scores. The z-score for $81 is (81 - 110.1) / 18 = -1.62, and for $142 it is (142 - 110.1) / 18 = 1.77.

We use the normalcdf function again, this time with -1.62 as the lower limit and 1.77 as the upper limit. The result is the proportion of bills between $81 and $142.

Part 3 of 3
(c) What is the probability that a randomly selected household had a monthly bill less than $110?

The z-score for $110 is (110 - 110.1) / 18 = -0.006. We use the normalcdf function with a large negative number as the lower limit (like -9999), and -0.006 as the upper limit. The result is the probability that a randomly selected household had a monthly bill less than $110.

Please note that the actual numbers may vary slightly due to rounding.

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