Question

Question 2 The average lifetime of a light bulb is 3,000 hours with a standard deviation of 696 hours A simple random sample of 36 bulbs is taken. a. What is the probability that the average life in the sample will be greater than 3,219 hours? What is the probability that the average life in the sample will be less than 3,180 hours? c. What is the probability that the average life in the sample will be between 2,670 and 2,809 hours?

Answer 1

SOLUTION X: lifetime of a random light bulb. aug lifestine is = 3000 hours → = E(X) = 3000 hours S.D. - Worlt) = 16 = 696 hours. A simple random sample x =(4, tq, . , X26) is token. (a) PC ***+ - + X36 > 3919) = ? 36 To solve the know the - ** above problem, we need to sampling distribution of Trs *96). 26 since n is lorge, le. n = 36 >30, by Col.T. Z: Z ~N10, 1). Pere jore, Flerefore, za 20 h ~ N10,1). 75 l 67336 3219-3000 - P(Z > - 696/236) =plz > 1.88)

No,۱) > 1.88 Plz osen under the curre right to 1.88 1.88 con be computed by using normal tuble. pl Z > 1088) = < 1.88) In pl Z [ P(Z >1.88)+ P(Z E 1.88)=17 -1 (11881 usere (3)=P(Z = z) = 1- 0.9699 = 0.0309 Arg. (b) P( 2 L 3180) 3180 . l = pl 7.6 lt L 6/6 6155) < 20 = P(Z 3150 - 3000 696) = P(Z L 1.55) - 0.9994 Any. N10 area under the normal curve left to 1.55. 1.55

For large na Z p- Elb) s. Elb) = p. P TPQ NNIO,1. P - 0.65 = Q = I-P & 1-0.65 -0.35 n = 200 PC & C 0.55) = 2 - T Pay , ( þ- p Jpg 0.55- P Pe/m 0.55 0.65 - Pr (Z L (0.65) (6.35) 200 - Pr (Z L - 2. 96) - v.001 Ans.