Question

Suppose that for a given computer salesperson, the probability distribution of x = the number of systems sold in one month is given by the following table.

x 1 2 3 4 5 6 7 8
p(x) .06 .10 .11 .30 .30 .11 .01 .01
(a) Find the mean value of x (the mean number of systems sold).
Mean = 4.1
(b) Find the variance and standard deviation of x. (Round the answers to four decimal places.)
Variance = 2.03
Standard deviation = 1.4248
(c) What is the probability that the number of systems sold is within 1 standard deviation of its mean value?
P (µ - s < x < µ + s ) = ?
(d) What is the probability that the number of systems sold is more than 2 standard deviations from the mean?
P ( x < µ - 2s or x > µ + 2s) = ?

Answer 1

(a) Mean = 4.1

(b) Variance = 2.03, Standard deviation = 1.4248

(c) P (µ - s < x < µ + s ) = 0.7451

(d) P ( x < µ - 2s or x > µ + 2s) = 0.022

Step-by-step explanation:

The mean (µ) of the probability distribution is calculated by summing the product of each value of x and its corresponding probability:

`\[ \mu = \sum_(i=1)^(8) x_i \cdot p(x_i) \]`

For the given distribution, the mean is (4.1).

The variance (σ²) is found by using the formula:

`\[ \sigma^2 = \sum_(i=1)^(8) (x_i - \mu)^2 \cdot p(x_i) \]`

After calculation, the variance is (2.03), and the standard deviation (σ) is the square root of the variance, giving (1.4248\).

To find the probability within 1 standard deviation of the mean (P (µ - s < x < µ + s)), we sum the probabilities of (x) between (3.6752) and (4.7248), resulting in (0.7451).

For the probability that the number of systems sold is more than 2 standard deviations from the mean (P ( x < µ - 2s or x > µ + 2s)), we calculate the sum of the probabilities for \(x\) less than (1.2504) or greater than (6.9496), yielding (0.022).

In summary, the mean, variance, and standard deviation provide key insights into the distribution, while probabilities within standard deviations offer a measure of the spread around the mean, and probabilities beyond 2 standard deviations indicate extreme values.

Answer 2

• A. The mean number of systems sold is 4.1.
• B.The standard deviation of x is approximately 1.4248.
• C. The probability that the number of systems sold is within 1 standard deviation of its mean value is 0.71.
• D. The probability that the number of systems sold is more than 2 standard deviations from the mean is 0.07.

Step-by-step explanation:

(a) To find the mean value of x (the mean number of systems sold), we multiply each value of x by its corresponding probability and sum up the results.

So, the mean is calculated as follows:

Mean =
`(1 * 0.06) + (2 * 0.10) + (3 * 0.11) + (4 * 0.30) + (5 * 0.30) + (6 * 0.11) + (7 * 0.01) + (8 * 0.01) = 4.1`

Therefore, the mean number of systems sold is 4.1.

(b) To find the variance and standard deviation of x, we need to calculate the variance first. The variance is a measure of how spread out the values are from the mean.

Variance =
`( (1 - 4.1)^2 * 0.06 ) + ( (2 - 4.1)^2 * 0.10 ) + ( (3 - 4.1)^2 * 0.11 ) + ( (4 - 4.1)^2 * 0.30 ) + ( (5 - 4.1)^2 * 0.30 ) + ( (6 - 4.1)^2 * 0.11 ) + ( (7 - 4.1)^2 * 0.01 ) + ( (8 - 4.1)^2 * 0.01 ) = 2.03`

The variance of x is 2.03.

Next, we can find the standard deviation by taking the square root of the variance.

Standard deviation = √(2.03) ≈ 1.4248

Therefore, the standard deviation of x is approximately 1.4248.

(c) To find the probability that the number of systems sold is within 1 standard deviation of its mean value, we need to calculate the range within which the number of systems sold falls.

The range within 1 standard deviation of the mean is from µ - s to µ + s, where µ is the mean and s is the standard deviation.

In this case, µ = 4.1 and s ≈ 1.4248.

So, the probability is given by:

P(µ - s < x < µ + s) = P(4.1 - 1.4248 < x < 4.1 + 1.4248) = P(2.6752 < x < 5.5248)

We can sum up the probabilities of x = 3, 4, and 5 to find the probability within 1 standard deviation.

P(2.6752 < x < 5.5248) = 0.11 + 0.30 + 0.30 = 0.71

Therefore, the probability that the number of systems sold is within 1 standard deviation of its mean value is 0.71.

(d) To find the probability that the number of systems sold is more than 2 standard deviations from the mean, we need to calculate the range beyond which the number of systems sold falls.

The range beyond 2 standard deviations of the mean is x < µ - 2s or x > µ + 2s, where µ is the mean and s is the standard deviation.

In this case, µ = 4.1 and s ≈ 1.4248.

So, the probability is given by:

P(x < µ - 2s or x > µ + 2s) = P(x < 4.1 - 2 * 1.4248 or x > 4.1 + 2 * 1.4248) = P(x < 1.2504 or x > 6.9496)

We can sum up the probabilities of x = 1 and x = 8 to find the probability beyond 2 standard deviations.

P(x < 1.2504 or x > 6.9496) = 0.06 + 0.01 = 0.07

Therefore, the probability that the number of systems sold is more than 2 standard deviations from the mean is 0.07.