The probability distribution for the number of defects in a shipment of alarm clocks, based on past data, is given below. Find the expected number of defects in a shipment of alarm clocks. Number of defects, n: 0, 1, 2, 3, 4Probability of n defects: 0.82, 0.11, 0.04, 0.02, 0.01

0.29

Step-by-step explanation:

Given :

n : ___ 0 1 2 3 4

P(n) : 0.82___ 0.11 ___0.04 _0.02 ___0.01

The expected number of defect in a shipment can be obtained using the expected value formula :

Expected value, E(X) = Σx*p(x)

Σx*p(x) = (0*0.82) + (1*0.11) + (2*0.04) + (3*0.02) + (4*0.01)

E(X) = 0.29

Hence, the expected number of defect in shipment is 0.29

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

one more happy customer!

< 3

step-by-step explanation:

i don't know, but that looks pretty bad. my school passed us unless we failed in all the subjects, but idk abt ur school's rules.

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