The distance travelled in a week by sales staff employed by a nationwide security firm is approximately normally distributed with a mean of 200 miles and standard deviation of 48 miles.
a) less than 80 miles,
b) more than 250 miles,
c) between 100 and 300 miles.
d) A reduced mileage rate is to be introduced which is to be set such that approximately 20% of the sales staff will be affected? What is the maximum number of miles per week that a member of staff could travel before reaching the reduced mileage rate?
e) Complete this sentence:
”80% of sales staff travel between ---------------------------- and ---------------------------- miles each week”
Can someone help?
a-c) Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like like "areas under normal distribution" to find the proportion/probability related to the Z scores.
(d,e) Look up % in table to get Z score.
Of course! I can help you with these questions.
a) To find the probability that a sales staff member will travel less than 80 miles in a week, we need to calculate the cumulative probability up to 80 miles. We can use the Z-score formula as follows:
Z = (X - μ) / σ
Where X is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.
Substituting the given values, we have:
Z = (80 - 200) / 48
Calculating this value, we find:
Z ≈ -2.5
Now, we need to find the cumulative probability corresponding to this Z-score using a standard normal distribution table or a statistical calculator. The cumulative probability for a Z-score of -2.5 is approximately 0.0062.
So, the probability that a sales staff member will travel less than 80 miles in a week is approximately 0.0062 or 0.62%.
b) To find the probability that a sales staff member will travel more than 250 miles in a week, we follow a similar process. The Z-score formula remains the same:
Z = (X - μ) / σ
Substituting the given values, we have:
Z = (250 - 200) / 48
Calculating this value, we find:
Z ≈ 1.04
To find the probability of a value greater than 1.04, we subtract the cumulative probability from the total probability of 1. The cumulative probability for a Z-score of 1.04 is approximately 0.8515.
So, the probability that a sales staff member will travel more than 250 miles in a week is approximately (1 - 0.8515) = 0.1485 or 14.85%.
c) To find the probability that a sales staff member will travel between 100 and 300 miles in a week, we need to find the cumulative probability for both values separately and subtract the two probabilities.
For 100 miles:
Z = (100 - 200) / 48
Z ≈ -2.08
The cumulative probability for a Z-score of -2.08 is approximately 0.0192.
For 300 miles:
Z = (300 - 200) / 48
Z ≈ 2.08
The cumulative probability for a Z-score of 2.08 is approximately 0.9808.
Thus, the probability that a sales staff member will travel between 100 and 300 miles in a week is approximately (0.9808 - 0.0192) = 0.9616 or 96.16%.
d) To find the maximum number of miles per week a member of staff could travel before reaching the reduced mileage rate (a rate that affects approximately 20% of the sales staff), we need to find the Z-score that corresponds to the 20th percentile.
Using a standard normal distribution table or calculator, we find that the Z-score corresponding to the 20th percentile is approximately -0.84.
Now, we can use the Z-score formula to find the maximum number of miles:
Z = (X - μ) / σ
Substituting the given values and rearranging the formula, we have:
X = Z * σ + μ
X = -0.84 * 48 + 200
Calculating this value, we find:
X ≈ 162.72
So, the maximum number of miles per week that a member of staff could travel before reaching the reduced mileage rate is approximately 162.72 miles.
e) To complete the sentence "80% of sales staff travel between ---------------------------- and ---------------------------- miles each week," we need to find the Z-scores corresponding to the 10th and 90th percentiles.
The Z-score corresponding to the 10th percentile is approximately -1.28, and the Z-score corresponding to the 90th percentile is approximately 1.28.
Now, we can use the Z-score formula to find the corresponding distances:
For the lower bound:
X1 = Z1 * σ + μ
X1 = -1.28 * 48 + 200
Calculating this value, we find:
X1 ≈ 139.84
For the upper bound:
X2 = Z2 * σ + μ
X2 = 1.28 * 48 + 200
Calculating this value, we find:
X2 ≈ 260.16
So, "80% of sales staff travel between 139.84 and 260.16 miles each week."