Download Quiz 5 Solutions

Quiz 5 Solutions
May 11, 2011
1
Suppose that 90% of all batteries from a certain supplier have acceptable voltages. A certain type of flashlight requires two type-D batteries, and the flashlight
will only work if both its batteries have acceptable voltages. Among ten randomly selected flashlights, what is the probability that at least nine will work?
The probability that a given flashlight will work is just the probability that
both its batteries have acceptable voltages, and presumably the batteries are
independent, so this probability is p = .9 × .9 = .81. Then the probability of k
of the ten randomly selected flashlights working is just given by B(k, 10, .81).
Since we want AT LEAST nine of the flashlights to work we need to sum over
the cases k = 9 and k = 10; that is
10
10
9
P (at least 9 will work) =
(.81) (.19) +
(.81)1 0
9
10
= 10(.81)9 (.19) + (.81)1 0
≈ .407
2
Let X Bin(n, p) be a binomial random variable.
2.1
For fixed n, are there values of p(0 ≤ p ≤ 1), for which V (X) = 0? Explain why
this is so.
The variance of a binomially distributed random variable is given by np(1 − p),
which is 0 if and only if p = 0 or p = 1. This corresponds to events which never
or always happen, respectively.
1
2.2
For what value of p is V (X) maximized?
We use the formula to write the variance as a funtion of p.
V (p) = np(1 − p) = n(p − p2 )
V 0 (p) = n(1 − 2p) = 0
if and only if p = 21 . The second derivative test or noticing that this is a
downwards-opening parabola confirms that this does indeed give a maximum.
2