The accompanying table describes results of roadworthiness tests of cars that ar
ID: 3204944 • Letter: T
Question
The accompanying table describes results of roadworthiness tests of cars that are 3 years old. The random variable x represents the numbers of cars that failed among six that were tested for roadworthiness. Find the mean and standard deviation for the number of cars that failed among the six cars that are tested.
Probabilities of Numbers of Cars that Failed among Six Tested
x (Number of Cars that Failed)
P(x)
0
0.3070.307
1
0.2660.266
2
0.2430.243
3
0.1110.111
4
0.0730.073
5
0+
6
0+
Probabilities of Numbers of Cars that Failed among Six Tested
x (Number of Cars that Failed)
P(x)
0
0.3070.307
1
0.2660.266
2
0.2430.243
3
0.1110.111
4
0.0730.073
5
0+
6
0+
Explanation / Answer
In order to find the mean (expectation) of the given distribution, we will use the following formula:
=xp(x)
So, first we need to multiply each value of X by each probability P(X), then add these results together. In this example we have:
=0307/1000+1133/500+2243/1000+3111/1000+473/1000=1377/1000
Now we will find the sum:
x2p(x)=02307/1000+1^2133/500+2^2243/1000+3^2111/1000+4^273/1000=681200
Putting all together we have:
=sqrt(x2p(x)2)=sqrt(681200(1377/1000)2)1.2284
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.