The designer of a cockpit of a new aircraft wants to position a switch so that m
ID: 3200221 • Letter: T
Question
The designer of a cockpit of a new aircraft wants to position a switch so that most pilots can reach it without having to change position. Suppose that the distribution among airline pilots of the maximum distance (measured from the back of the seat) that can be reached without moving the seat is approximately Normally distributed with a mean µ =125cm and a standard deviation of 10 cm.
a) If the switch is placed 120cm from the back of the seat, what proportion of pilots will be able to reach it without moving the seat?
b) What is the maximum distance from the back of the seat that the switch could be placed if it is required that 95% of pilots will be able to reach it without moving the seat?
c) If a pilot has a z-score of 1.5, what does this mean in context? To what maximum does a zscore of 1.5 correspond?
Explanation / Answer
Answer to part a)
We have Mean (M) = 125 cm
Standard deviation (S) = 10
The switch is placed at X = 120 cm
.
The formula of Z is:
z = (X - M) / S
z = (120 -125)/10 = -0.5
.
P(Z<0.5) can be obtained from the Z table:
We get P(Z < -0.5) = 0.3085
.
Answer to part b)
if we want 95% of the pilots to reach the switch without moving from their seats, we need to find the value of x for it
we have:
Mean (M) = 125
Standard deviation (S) = 10
P = 0.95
.
Thus we can find the value of Z corresponding to P = 0.95 from the Z table
We get Z = 1.645
.
Thus we can make use of the Z formula
1.645 = (X - 125)/10
16.45 = X - 125
X = 16.45 + 125 = 141.45
Thus the maximum distance from the back of the seat can be 141.45 cm
.
Answer to part c)
A Z score of 1.5 indicates that the seat of the pilot is 1.5 times the standard deviation above the mean.
It would correspond to 1.5 * 10 + 125 = 140 cms