Question #270436

The manufacturer of a certain type of light bulb believes that the average life of a light bulb is approximately 1000 hours with a standard deviation of 100 hours.

(a) If a light bulb is selected at random from a huge consignment, what is the

probability that it will last for:

Less than 1700 hours?

More than 1780 hours?

Between 1700 hours and 1780 hours?

(b) Above what life time (in hours) would the longest 5% working light bulbs

last? (c) If the manufacturer in (a) above, supplies 3000 light bulbs, how many

would be expected to last between 1700 hours and 1780 hours


1
Expert's answer
2021-11-29T05:06:55-0500

μ=1000σ=100\mu=1000 \\ \sigma = 100

(a)

P(X<1700)=P(Z<17001000100)=P(Z<7)=0.999968P(X>1780)=1P(X<1780)=1P(Z<17801000100)=1P(Z<7.8)=10.999968=0.000032P(1700<X<1780)=P(X<1780)P(X<1700)=P(Z<17801000100)P(Z<17001000100)=P(Z<7.8)P(Z<7)=0.9999680.999968=0P(X<1700) = P(Z< \frac{1700-1000}{100}) \\ = P(Z< 7) \\ = 0.999968 \\ P(X>1780) = 1 -P(X<1780) \\ = 1 -P(Z< \frac{1780-1000}{100}) \\ = 1 -P(Z< 7.8) \\ = 1-0.999968 \\ = 0.000032 \\ P(1700<X<1780) = P(X< 1780) -P(X<1700) \\ = P(Z< \frac{1780-1000}{100}) -P(Z< \frac{1700 -1000}{100}) \\ = P(Z< 7.8) -P(Z<7) \\ = 0.999968 -0.999968 \\ = 0

(b) Let the variable A denote life time in hours that would the longest 5% working light bulb last, and calculation for A is given below.

A5=μZ0.05σA=1000(1.645)(100)A=1164.5A_5 = \mu -Z_{0.05} \sigma \\ A = 1000 -(-1.645)(100) \\ A = 1164.5

(c) N=3000

The required number =3000×0=0= 3000 \times 0 = 0


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