1. A mail-order computer business has six telephone lines. Let X ...

13. The lifetime of a certain type of battery is normally distributed with
mean value 8 hours and standard deviation 1 hour. There are four
batteries in a package. What lifetime value is such that the total
lifetime of all batteries in a package exceeds that value for only 5% of
all packages?
Ans
Individual lifetimes be denoted by Xi ∼ N(µ, σ 2 ). Then the total
lifetime of 4 batteries is given by Y = X1 + X2 + X3 + X4. Therefore,
Y ∼ N(4µ, 4σ 2 ) or Y ∼ N(4 × 8, 4 × 1 2 ) or Y ∼ N(32, 4). Therefore,
[Y − 32]/2 ∼ N(0, 1). Let z0 be the required critical point, then
z0 − 32
Φ
= 1 − 0.05 = 0.95
2
We know that Φ(1.644854) = 0.95. Therefore,
z0 − 32
2
= 1.644854 ⇒ z0 = 32 + 2 × 1.644854 = 35.28971
14. Each of 150 newly manufactured items is examined and the number
of scratches per item is recorded (the items are supposed to be free of
scratches), yielding the following data:
Scratches/item 0 1 2 3 4 5 6 7
Observed freq 18 37 42 30 13 7 2 1
Let X = the number of scratches on a randomly selected item and
assume that X has a Poisson distribution with parameter λ (P (X =
k) = exp(−λ)λ k /k!).
Find an unbiased estimator of λ and compute the estimate for the
above data. [Hint: E(X) = λ)
Ans If E[X] = λ, implies
E[ ¯ X = E[ 1
n
n
i=1
Xi] = 1
n
= [nλ]/n = λ
n
E[Xi]
Therefore ¯ X is an unbiased estimator of λ. The estimate is simply,
the mean of the above data,
= (0 × 18 + 1 × 37 + 2 × 42 + 3 × 30 + 4 × 13 + 5 × 7 + 6 × 2 + 7 × 1)
(18 + 37 + 42 + 30 + 13 + 7 + 2 + 1)
= 2.113333
9
i=1