Download Stat-152 Homework #4

Stat-152 Homework #5
Solutions
4.3
a) In order to find the probability of exactly 3 defects in the new car, P(X=3), we will make use of the fact
that if we add up the probability of all possible outcomes, the sum is equal to 1. Symbolically,
1 = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)=> 1 = .5 + .3 + .1 + P(X=3) + .05
=> 1 = .95 + P(X=3). Solving this equation for P(X=3), we get P(X=3) = .05.
b) There are two ways of finding the probability that there are at most 2 defects in a new car: P(X ≤2).
The first method is to note that P(X ≤2) = P(X=0) + P(X=1) +P(X=2) = .5 + .3 + .1 = . 9
The other method is to note that P(X ≤2)= 1 – P(X > 2) = 1 – P(X=3) + P(X=4) = 1 - .1 = . 9
Regardless of which method you use, you find that P(X ≤2) = . 9
c) We need to find the mean and standard deviation of the defects.
E[X] = 0(.5) + 1(.3) + 2(.1) + 3(.05) + 4(.05) = .85
To find σ, we first need to find σ2 = Var(X).
Var(X) = (0 - .85)2(.5) + (1 - .85)2(.3) + (2 - .85)2(.1) + (3 - .85)2(.05) + (4 - .85)2(.05) = 1.2275.
Then σ = √(Var(X)) = √(1.2275) = 1.1079
4.7
a) E[X] = 0(.998) + 100(.001) + 1000(.001) = 1.1
b) Var(X) = (0 – 1.1)2(.998) + (100 – 1.1)2(.001) + (1000 – 1.1)2(.001) = 1008.79
c) If you pay $2.00 for a ticket, and your expected winnings are $1.1, then it doesn’t make much sense to
play the lottery. Even though there is a lot of variability to the data, your odds of winning are
astronomically small.
4.9
a) To find the probability that you don’t get your license in the first two attempts, we can either add up the
probability of getting your license in 3,4, and 5 attempts. Note that P(X=3) = 1-.42-.33-.04 -.01 = .2
So P(Not getting in the first two) = P(X=3) + P(X=4) + P(X=5) = .2 +.04+.01 = . 25
Similarly, we can also find this probability by taking 1- P(Getting the license in the first two attempts)
= 1 – P(X=1) – P(X=2) = 1 - .42 -.33 => 1 - .75 = . 25. So regardless of how you go about it, we find the
probability of not passing your test in the first two attempts is . 25
b) To find the mean and standard deviation, first we find the mean.
E[X] = 1(.42) + 2(.33) + 3(.2) + 4(.04) + 5(.01) = 1.89
Var(X) = (1 – 1.89)2(.42) + (2 – 1.89)2(.33) + (3 – 1.89)2(.2) + (4 – 1.89)2(.04) + (5 – 1.89)2(.01)
So Var(X) = .8579.
Therefore, the standard deviation = √(.8579) = .9262
4.13
a) The phrase “the density curve is constant” implies two things: 1) This distribution is continuous, 2) the
density curve looks as follows:
So in order to find the height of the curve, we use the fact that the area of the rectangle is equal to 1, and
that the length of the rectangle is 8.
So 1 = 8 * height => height = 1/8.
b) It may be easier to view the following questions graphically
i) P(Dismisses before 3 minutes)
So the probability is equal to the area under the rectangle = (1/8)*3 = 3/8
ii) P(Between 5 and 6 minutes)
Again, the probability is equal to the area under the rectangle = (1/8)*1 = 1/8
c) The mean of the length of time the professor lectures over is equal to the midpoint of this density curve.
In this specific case, it’s the half way point on the graph, i.e. when X=4. So E[X]=4. This can be verified
using calculus, but for the sake of this class, the above definition will do just fine. As far as the standard
deviation, it is positive and probably somewhere in between half the distance between the mean and the
endpoints. Again, using calculus, this can be found to be about 2.33