Question
Jan Villaroel
Topic: Statistics Posted 11 months ago
A business has six customer service telephone lines. Consider the random variable x = Number of lines in use at a randomly selected time.

A business has six customer service telephone lines. Consider the random variable
x = Number of lines in use at a randomly selected time. Suppose that the probability distribution of x is as
follows

x 0 1 2 3 4 5 6
p(x) 0.05 0.10 0.20 0.35 0.20 0.08 0.02

(a) Calculate the mean value and standard deviation of x. (Round your standard deviation to four decimal
places.)

(b) What is the probability that the number of lines in use is farther than 3 standard deviations from the
mean value?


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Koala Tutor
Jan 6, 2024
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(a) To calculate the mean value of x, we multiply each value of x by its corresponding probability and sum them up:

Mean (μ) = (0 * 0.05) + (1 * 0.10) + (2 * 0.20) + (3 * 0.35) + (4 * 0.20) + (5 * 0.08) + (6 * 0.02)
= 0 + 0.10 + 0.40 + 1.05 + 0.80 + 0.40 + 0.12
= 2.87

So, the mean value of x is 2.87.

To calculate the standard deviation, we first need to calculate the variance. The variance (σ^2) is calculated by summing up the squared differences between each value of x and the mean, multiplied by their probabilities:

Variance (σ^2) = [(0 - 2.87)^2 * 0.05] + [(1 - 2.87)^2 * 0.10] + [(2 - 2.87)^2 * 0.20] + [(3 - 2.87)^2 * 0.35] + [(4 - 2.87)^2 * 0.20] + [(5 - 2.87)^2 * 0.08] + [(6 - 2.87)^2 * 0.02]
= [(-2.87)^2 * 0.05] + [(-1.87)^2 * 0.10] + [(-0.87)^2 * 0.20] + [(-0.13)^2 * 0.35] + [(1.13)^2 * 0.20] + [(2.13)^2 * 0.08] + [(3.13)^2 * 0.02]
= [8.2565 * 0.05] + [3.5009 * 0.10] + [0.6749 * 0.20] + [0.0221 * 0.35] + [1.2777 * 0.20] + [1.8168 * 0.08] + [3.5998 * 0.02]
= 0.4128 + 0.3501 + 0.13498 + 0.007735 + 0.25554 + 0.14534 + 0.071996
= 1.3784

To calculate the standard deviation (σ), we take the square root of the variance:

Standard Deviation (σ) = √1.3784
= 1.1738

So, the standard deviation of x is approximately 1.1738 (rounded to four decimal places).

(b) For a normal distribution, approximately 99.73% of the data falls within 3 standard deviations from the mean. Therefore, the probability that the number of lines in use is farther than 3 standard deviations from the mean is approximately 0.27%.

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