Answer to Question #254187 in Statistics and Probability for Anne

Question

1.Light bulbs having a mean life of 2400 hours and a standard deviation of 62 hours are used for a consignment of 4200 bulbs.

    a.) Determine the number of bulbs likely to have a life in excess of 2500 hours.

    b.) Determine the percentage of bulbs with life in between 2200 hours and 2500 hours.

    c.) Determine the percentage of bulbs with life in between 2400 hours and 2600 hours

Answer 1

$\mu = 2400 \\ \sigma = 62 \\ N= 4200$

a)

$P(X>2500) = 1 -P(X<2500) \\ = 1 -P(Z< \frac{2500-2400}{62}) \\ = 1 -P(Z< 1.613) \\ = 1 -0.9466 \\ =0.0534 \\ Needed \; number = 0.0534 \times 4200 = 224.28 ≈ 224$

b)

$P(2200<X<2500) = P(X<2500) -P(X<2200) \\ = P(Z< \frac{2500-2400}{62}) -P(Z< \frac{2200-2400}{62}) \\ = P(Z< 1.613) -P(Z< -3.225) \\ = 0.9466 -0.00063 \\ = 0.94597 \\ = 94.597 \; \%$

c)

$P(2400<X<2600) = P(X<2600) -P(X<2400) \\ = P(Z< \frac{2600-2400}{62}) -P(Z< \frac{2400-2400}{62}) \\ = P(Z< 3.225) -P(Z< 0) \\ = 0.99937 -0.5 \\ = 0.49937 \\ = 49.937 \; \%$