The probabilities of a machine manufacturing 0, 1, 2, 3,

Ikunupe6v

Ikunupe6v

Answered question

2021-12-10

The probabilities of a machine manufacturing 0, 1, 2, 3, 4 or 5 defective parts in one day are 0.75, p, 0.04, 0.025, 0.01 and 0.005 respectively. Construct a probability distribution for the data, by finding the missing probability. Also find the mean and variance.

Answer & Explanation

yotaniwc

yotaniwc

Beginner2021-12-11Added 34 answers

Step 1 
Total probability is always 1. So 
0.75+p+0.04+0.025+0.01+0.005=1 
0.83+p=1 
p=10.83 
p=0.17 
Total probability is consistently 1. So
number of defective parts (xi)012345probability(pi)0.750.170.040.0250.010.005 
Step 2 
Mean=E(X) 
=0(0.75)+1(0.17)+2(0.04)+3(0.025)+4(0.01)+5(0.005) 
=0.39 
E(X2) 
=02(0.75)+12(0.17)+22(0.04)+32(0.025)+42(0.01)+52(0.005) 
=0.84 
Variance 
=E(X2)[E(X)]2 
=0.840.392 
=0.6879 
Result: Mean=0.39 
Variance =0.6879

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