In: Statistics and Probability
their respective probabilities as shown below.
Number of |
5 .01 |
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A.) The expected number of machine breakdowns per month
B.) The variance of machine breakdowns per month
C.) The standard deviation of machine breakdowns per month
Solution :
= 0 *0.15 + 1 *0.41 + 2 * 0.22 + 3 * 0.17+ 4 * 0.04+5*0.01
= ( 0+ 0.41 + 0.44 + 0.51+0.16+0.05)
= 1.57
(B)
= [ 02 *0.15 + 12 *0.41 + 22 * 0.22 + 32 * 0.17+ 42 * 0.04+52 *0.01]-1.57
= [( 0+0.41 + 0.88 + 1.53+0.64+0.25) )] -2.4649
= 3.71 -2.4649
= [
02 *0.15 + 12 *0.41 + 22 * 0.22 + 32 * 0.17+ 42 * 0.04+52
*0.01]-1.57
=
[( 0+0.41 + 0.88 + 1.53+0.64+0.25) )] -2.4649