Solved: "The probabilities of a machine manufacturing 0, 1,..."

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

Beginner2021-12-11Added 34 answers

Step 1
Total probability is always 1. So
$0.75 + p + 0.04 + 0.025 + 0.01 + 0.005 = 1$
$0.83 + p = 1$
$p = 1 - 0.83$
$p = 0.17$
Total probability is consistently 1. So
$\begin{array}{cc} number of defective parts (x_{i}) & 0 & 1 & 2 & 3 & 4 & 5 \\ probability (p_{i}) & 0.75 & 0.17 & 0.04 & 0.025 & 0.01 & 0.005 \end{array}$
Step 2
$M e a n = E (X)$
$= 0 (0.75) + 1 (0.17) + 2 (0.04) + 3 (0.025) + 4 (0.01) + 5 (0.005)$
$= 0.39$
$E (X^{2})$
$= 0^{2} (0.75) + 1^{2} (0.17) + 2^{2} (0.04) + 3^{2} (0.025) + 4^{2} (0.01) + 5^{2} (0.005)$
$= 0.84$
Variance
$= E (X^{2}) - {[E (X)]}^{2}$
$= 0.84 - {0.39}^{2}$
$= 0.6879$
Result: $M e a n = 0.39$
Variance $= 0.6879$

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