A sample of 5 items is selected at random fro. a box containing 15 items of which 8 are defective find 1) Mean 2) variance of defective items
Answers
We have a lot of 6 items, 2 are defective →6−2=4 are non-defective.
P (drawing a defective item) =
6
2
=
3
1
P (drawing a non-defective item) = 1 - P (drawing a defective item) =1−
3
1
=
3
2
Note:4 items can be drawn from a lot of 6 items in
6
C
4
ways.
Let X be the random variable of drawing 2 defective items from the lot.
P(X=0)= P (no defective item in the sample) =
6
C
4
4
C
4
=
15
1
P(X=1)= P (one defective item from the lot) = P(one defective, 3 non-defective items) =(
6
C
4
2
C
1
4
C
3
)=
15
8
P(X=2)= P (two defective items from the lot) = P(2 defective, 2 non-defective bulbs) =
6
C
4
2
C
2
4
C
2
=
15
6
Therefore the required probability distribution is as follows:
X: 0 1 2
P(X):
15
1
15
8
15
6
So Mean of X will be:∑
i=0
2
x
i
P(X=x
i
)
→0×
15
1
+1×
15
8
+2×
15
6
=
3
4