Question

A popular flashlight that uses two D-size batteries was selected, and several of the same models were purchased to test the "continuous-use life" of D batteries. As fresh batteries were installed, each flashlight was turned on and the time noted. When the flashlight no longer produced light, the time was again noted. The resulting "life" data from Rayovac batteries had a mean of 21.8 hours. Assume these values have a normal distribution with a standard deviation of 1.38 hours. (Give your answers correct to four decimal places.)

(a) What is the probability that one randomly selected Rayovac battery will have a test life between 20.1 and 22.3 hours?

(b) What is the probability that a randomly selected sample of 5 Rayovac batteries will have a mean test life between 20.1 and 22.3 hours?

(c) What is the probability that a randomly selected sample of 17 Rayovac batteries will have a mean test life between 20.1 and 22.3 hours?

(d) What is the probability that a randomly selected sample of 62 Rayovac batteries will have a mean test life between 20.1 and 22.3 hours?

(e) Describe the effect that the increase in sample size had on the answers for parts (b) - (d). As sample size increased, the standard error increased, resulting in higher probabilities. As sample size increased, the standard error increased, resulting in lower probabilities. As sample size increased, the standard error decreased, resulting in higher probabilities. As sample size increased, the standard error decreased, resulting in lower probabilities.

Answer 1

a)

b)

Given mean = 21.8

Standard deviation =1.38

sample n = 5

P(20.1 < X < 22.3) = P(  x - μ/ σ / √n < z <  x - μ/ σ / √n)

= P( 20.1 - 21.8 / 1.38 / √5 < z < 22.8 - 21.8 / 1.38 / √5 )

= P( -2.75 < z < 1.62)

P (−0.75<Z<1.62 ) = 0.7208

c)

Given mean = 21.8

Standard deviation =1.38

sample n = 5

P(20.1 < X < 22.3) = P(  x - μ/ σ / √n < z <  x - μ/ σ / √n)

= P( 20.1 - 21.8 / 1.38 / √17 < z < 22.8 - 21.8 / 1.38 / √17 )

= P( -5.079 < z < 2.98)

P (−5.079<Z<2.98 )=0.9986

d)

Given mean = 21.8

Standard deviation =1.38

sample n = 5

P(20.1 < X < 22.3) = P(  x - μ/ σ / √n < z <  x - μ/ σ / √n)

= P( 20.1 - 21.8 / 1.38 / √62 < z < 22.8 - 21.8 / 1.38 / √62 )

= P( -9.69 < z < 5.70)

P (−9.69<Z<5.70 )=1

*** Dear student we answer Four sub parts per question post remaining separately****