Answer:
a) The minimum head breadth that will fit the clientele = 4.105 inches to 3d.p = 4.1 inches to 1 d.p
b) The maximum head breadth that will fit the clientele = 8.905 inches to 3 d.p = 8.9 inches to 1 d.p
Step-by-step explanation:
This is normal distribution problem.
A normal distribution has all the data points symmetrically distributed around the mean in a bell shape.
For this question, mean = xbar = 6.1 inches
Standard deviation = σ = 1 inch
And we want to find the lowermost 2.3% and uppermost 2.3% of the data distribution.
The minimum head breadth that will fit the clientele has a z-score with probability of 2.3% = 0.023
Let that z-score be z'
That is, P(z ≤ z') = 0.023
Using the table to obtain the value of z'
z' = - 1.995
P(z ≤ - 1.995) = 0.023
But z-score is for any value, x, is that value minus the mean then divided by the standard deviation.
z' = (x - xbar)/σ
- 1.995 = (x - 6.1)/1
x = -1.995 + 6.1 = 4.105 inches
The maximum head breadth that will fit the clientele has a z-score with probability of 2.3% also = 0.023
Let that z-score be z''
That is, P(z ≥ z'') = 0.023
Using the table to obtain the value of z''
P(z ≥ z") = P(z ≤ -z")
- z'' = - 1.995
z" = 1.995
P(z ≥ 1.995) = 0.023
But z-score is for any value, x, is that value minus the mean then divided by the standard deviation.
z'' = (x - xbar)/σ
1.995 = (x - 6.1)/1
x = 1.995 + 6.1 = 8.905 inches
