Tutorial 07 Solutions 1 2015
Tutorial 07: Solutions
Part A:
For all your answers, please remember to do the following:
1. Draw curves
2. State the distribution
3. Define the variable
A7.1
An automatic machine in a manufacturing process is operating properly if the lengths of an important subcomponent are normally distributed, with mean µ = 117 cm and standard deviation σ = 2.1 cm.
If the machine is operating correctly:
Let X = variable length of subcomponent (cm).
Then if the machine is operating correctly, X ~ N (117, 2.12 ) .
(i)
find the probability that one randomly selected unit has a length greater than 120 cm;
X −µ
120 − 117
) = P( Z > 1.4286) ≈ P( Z > 1.43) σ 2.1
= 1 − P( Z < 1.43) = 1 − 0.9236 = 0.0764
X:
P( X > 120) = P(
>
Alternative solution: P ( Z > 1.43) = P ( Z < −1.43) = 0.0764
(ii)
Z:
117
120
0
1.43
find the probability that if three units are randomly selected, their mean length exceeds 120 cm;
Let
X = mean length (cm), then
2 . 12
X ~ N 117 ,
3
P(X > 120) = P(
X − µ X 120 − 117
120 − 117
>
) = P(Z >
) = P(Z > 2.4744) σ 2.1
1.2124
n
3
X:
117
120
Z:
0
2.474
Using symmetry of curve, and then tables we find
P ( Z > 2.47) = P ( Z < −2.47) = 0.0068 ≈ 0.007
1
(iii)
explain the differences between parts (i) and (ii).
In (i) we are considering the probability that a single item is longer than 120cm.
The answer is about 0.076.
In (ii) we are considering three separate items, but the question is whether their mean length is greater than 120cm. This will certainly happen if all three are longer than 120cm, but it can also happen when one or two are shorter than
120, so long as the average is above. However the mean is less variable than individual values, so the probability of the mean being above 120cm will be lower than the probability for an individual item.
Part B:
B7.2
A statistical analyst who works for a large insurance company is in the process of examining several pension plans. Company records show that the age at which its male
Part A:
For all your answers, please remember to do the following:
1. Draw curves
2. State the distribution
3. Define the variable
A7.1
An automatic machine in a manufacturing process is operating properly if the lengths of an important subcomponent are normally distributed, with mean µ = 117 cm and standard deviation σ = 2.1 cm.
If the machine is operating correctly:
Let X = variable length of subcomponent (cm).
Then if the machine is operating correctly, X ~ N (117, 2.12 ) .
(i)
find the probability that one randomly selected unit has a length greater than 120 cm;
X −µ
120 − 117
) = P( Z > 1.4286) ≈ P( Z > 1.43) σ 2.1
= 1 − P( Z < 1.43) = 1 − 0.9236 = 0.0764
X:
P( X > 120) = P(
>
Alternative solution: P ( Z > 1.43) = P ( Z < −1.43) = 0.0764
(ii)
Z:
117
120
0
1.43
find the probability that if three units are randomly selected, their mean length exceeds 120 cm;
Let
X = mean length (cm), then
2 . 12
X ~ N 117 ,
3
P(X > 120) = P(
X − µ X 120 − 117
120 − 117
>
) = P(Z >
) = P(Z > 2.4744) σ 2.1
1.2124
n
3
X:
117
120
Z:
0
2.474
Using symmetry of curve, and then tables we find
P ( Z > 2.47) = P ( Z < −2.47) = 0.0068 ≈ 0.007
1
(iii)
explain the differences between parts (i) and (ii).
In (i) we are considering the probability that a single item is longer than 120cm.
The answer is about 0.076.
In (ii) we are considering three separate items, but the question is whether their mean length is greater than 120cm. This will certainly happen if all three are longer than 120cm, but it can also happen when one or two are shorter than
120, so long as the average is above. However the mean is less variable than individual values, so the probability of the mean being above 120cm will be lower than the probability for an individual item.
Part B:
B7.2
A statistical analyst who works for a large insurance company is in the process of examining several pension plans. Company records show that the age at which its male