Question

A company has discovered that a recent batch of batteries had manufacturing​ flaws, and has issued a recall. In a group of 15 batteries covered by the​ recall, 3 are dead. Two batteries at random are chosen from the package of 15

​b) Create a probability model for the number of good batteries chosen.

​c) What's the expected number of good​ batteries?

​d) What's the standard​ deviation?

Answer 1

Total batteries = 15, Bad = 3 and good = 15 - 3 = 12

For each case the total outcomes is choosing 2 batterries out of 15 = ¹⁵C₂ = 105

Break up of selection	X	P(x)
None from Good and 2 from Bad	0	(12C0 * 3C2) / 105 = (1 * 3)/105 = 3/105 = 1/35
1 from Good and 1 from bad	1	(12C1 * 3C1) / 105 = (12 * 3)/105 = 36/105 = 12/35
2 from Good and 0 from bad	2	(12C2 * 3C0) /105 = (66 * 1)/105 = 66/105 = 22/35

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(b) Therefore the probability distribution is

X	P(x)
0	1/35
1	12/35
2	22/35

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(c) Expected Value = E[x] = Sum[x * P(x)] = 0 * (1/35) + 1 * (12/35) + 2 * (22/35)

= 0 + 12/35 + 44/35 = 56/35 = 8/5 = 1.6.

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(d) Standard Deviation = SqrtVariance)

Variance = E[x²] - (E[x])² =

E[x²] = Sum[x² * P(x)] = 0² * (1/35) + 1² * (12/35) + 2² * (22/35)

= 0 + 12/35 + 88/35 = 100/35 = 20/7

(E[x])² = (8/5)² = 64/25

Therefore Variance = 20/7 - 64/25 = (500-448)/175 = 52/175

Therefore Standard Deviation = Sqrt(52/175) = 0.545

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