Question

The following is the probability distribution of the number of computer network interruptions in one day. x 0 1 2 3 4 5 P(X=x) 0.18 0.23 0.21 0.15 0.11 (a) Compute the value of 𝑎. (b) Find the probability that no computer network interruption is encountered in one day. (c) Find the probability of experiencing at least three computer network interruptions in one day. (d) Find the probability of experiencing at most four computer network interruptions in one day. (e) Find the expected value of 𝑋. (f) Find the variance and standard deviation of 𝑋.

          The following is the probability distribution of the number of computer network interruptions in one day.

x
0
1
2
3
4
5
P(X=x)
0.18
0.23
0.21
0.15
0.11

(a) Compute the value of 𝑎.
(b) Find the probability that no computer network interruption is encountered in one day.
(c) Find the probability of experiencing at least three computer network interruptions in one day.
(d) Find the probability of experiencing at most four computer network interruptions in one day.
(e) Find the expected value of 𝑋.
(f) Find the variance and standard deviation of 𝑋.
        
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Added by Willie B.

Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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The following is the probability distribution of the number of computer network interruptions in one day. x 0 1 2 3 4 5 P(X=x) 0.18 0.23 0.21 0.15 0.11 (a) Compute the value of 𝑎. (b) Find the probability that no computer network interruption is encountered in one day. (c) Find the probability of experiencing at least three computer network interruptions in one day. (d) Find the probability of experiencing at most four computer network interruptions in one day. (e) Find the expected value of 𝑋. (f) Find the variance and standard deviation of 𝑋.
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Transcript

-
00:01 Hello, let's have a look on the question.
00:02 So we know that summation of probability is always equal to 1.
00:05 So we can say that summation of all the probabilities, 0 .18 plus 0 .23 plus 0 .15, a and 0 .11 will be equal to 1.
00:19 So from here we can find out the value of a, that is 1 minus 0 .18, 0 .23, 0 .21, 0 .15, and 0 .15, and 0 .0 .0 .5, and 0.
00:30 So on calculating this value will be equal to 1 minus 0 .88, that is 0 .12.
00:37 So a is equal to 0 .12.
00:41 Now we need to compute probability x equal to 0.
00:45 So for the b part, probability x equal to 0, this will be equal to 0 .18.
00:55 For the c part, probability x is greater than equal to 3 is equal to probability that x equal to 3 x equal to 4 and probability x equal to 5 so this will be equal to 0 .15 plus 0 .12 plus 0 .11 that is 0 .38...
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