Distance travelled in a week by sales staff employed by nationwide security firm is approximately normally distributed with a mean of 200 miles and standard deviation of 48 miles. Estimate the percentage of staff who in a week travel:
1.
a)less than 80 miles
b)more than 250 miles
c)between 100 and 300 miles
2.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 miles per week that a member of staff could travel before reaching the reduced mileage rate?
3)
Complete this sentence:
"80% of sales staff travel between ____ and ______ miles each week".
I need help for this question asap thanx
1. z = (x - mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.
Reverse the process for the last two questions.
To answer these questions, we will use the information provided regarding the mean, standard deviation, and the assumption that the distance travelled follows a normal distribution.
1.
a) To estimate the percentage of staff who travel less than 80 miles in a week, we need to calculate the z-score and then find the corresponding area under the normal curve. The z-score is calculated by subtracting the mean from the given value and dividing by the standard deviation:
z = (x - μ) / σ
In this case, x = 80, μ = 200, and σ = 48. Plugging in these values, we get:
z = (80 - 200) / 48 = -2.5
Now, we need to find the percentage of the area under the normal curve corresponding to a z-score of -2.5. Using a standard normal distribution table or a calculator, we find that the area to the left of -2.5 is approximately 0.0062. This means that about 0.62% of the staff travel less than 80 miles in a week.
b) To estimate the percentage of staff who travel more than 250 miles in a week, we follow a similar process. The z-score is calculated as:
z = (x - μ) / σ
In this case, x = 250, μ = 200, and σ = 48. Plugging in these values, we get:
z = (250 - 200) / 48 = 1.04
Using a standard normal distribution table or a calculator, we find that the area to the right of 1.04 is approximately 0.1492. This means that about 14.92% of the staff travel more than 250 miles in a week.
c) To estimate the percentage of staff who travel between 100 and 300 miles in a week, we need to find the area under the normal curve between two z-scores. We can calculate the z-scores as follows:
z1 = (x1 - μ) / σ = (100 - 200) / 48 = -2.08
z2 = (x2 - μ) / σ = (300 - 200) / 48 = 2.08
Using a standard normal distribution table or a calculator, we find that the area between -2.08 and 2.08 is approximately 0.9798. This means that about 97.98% of the staff travel between 100 and 300 miles in a week.
2.
To find the maximum miles per week that a member of staff could travel before reaching the reduced mileage rate that affects approximately 20% of the sales staff, we need to find the z-score corresponding to the desired percentage. In this case, the desired percentage is 80%, which corresponds to an area of 0.8 under the normal curve.
Using a standard normal distribution table or a calculator, we find that the z-score corresponding to an area of 0.8 is approximately 0.8416. We can then use the z-score formula to find the maximum miles per week (x) as follows:
0.8416 = (x - μ) / σ
Substituting the given values, μ = 200 and σ = 48, we can solve for x:
0.8416 = (x - 200) / 48
x - 200 = 48 * 0.8416
x = 200 + (48 * 0.8416)
x ≈ 239.2
Therefore, a member of staff could travel up to approximately 239.2 miles per week before reaching the reduced mileage rate.
3)
"80% of sales staff travel between ____ and ______ miles each week".
From the previous answer, we know that a member of staff could travel up to approximately 239.2 miles per week before reaching the reduced mileage rate. Therefore, we can complete the sentence as follows:
"80% of sales staff travel between 0 and 239.2 miles each week."