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 20.9 hours. Assume these values have a normal distribution with a standard deviation of 1.47 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 19.9 and 21.3 hours? 0.359 Correct: Your answer is correct.

(b) What is the probability that a randomly selected sample of 6 Rayovac batteries will have a mean test life between 19.9 and 21.3 hours? 0.1502 Incorrect: Your answer is incorrect. seenKey 0.6996

(c) What is the probability that a randomly selected sample of 12 Rayovac batteries will have a mean test life between 19.9 and 21.3 hours? _________.

(d) What is the probability that a randomly selected sample of 65 Rayovac batteries will have a mean test life between 19.9 and 21.3 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)

= 20.9

= 1.47

To find P(19.9 < X < 21.3)

Case 1: For X from 19.9 to mid value:
Z = (19.9 - 20.9)/1.47

= - 0.6803

Technology gives area = 0.2518

Case 2: For X from mid value to 21.3:
Z = (21.3 - 20.9)/1.47

= 0.2721

Technology gives area = 0.1072

So,

P(19.9 < X < 21.3) = 0.2518 + 0.1072 = 0.3590

So,

Answer is:

0.3590

(b)

n = 6

SE = /

= 1.47/

= 0.6001

To find P(19.9 < < 21.3):

Case 1: For from 19.9 to mid value:

Z = (19.9 - 20.9)/0.6001

= - 1.6663

Technology gives area = 0.4522

Case 2: for from mid value to 21.3:

Z = (21.3 - 20.9)/0.6001

= 0.6667

Technology gives area = 0.2475

So,

P(19.9 < < 21.3) = 0.4521 + 0.2475 = 0.6996

So,

Answer is:

0.6996

(c)

n = 12

SE = /

= 1.47/

= 0.4244

To find P(19.9 < < 21.3):

Case 1: For from 19.9 to mid value:

Z = (19.9 - 20.9)/0.4244

= - 2.3565

Technology gives area = 0.4908

Case 2: for from mid value to 21.3:

Z = (21.3 - 20.9)/0.4244

= 0.9425

Technology gives area = 0.3270

So,

P(19.9 < < 21.3) = 0.4908 + 0.3270 = 0.8178

So,

Answer is:

0.8178

(d)

n = 65

SE = /

= 1.47/

= 0.1823

To find P(19.9 < < 21.3):

Case 1: For from 19.9 to mid value:

Z = (19.9 - 20.9)/0.1823

= - 5.4845

Technology gives area = 0.5 nearly

Case 2: for from mid value to 21.3:

Z = (21.3 - 20.9)/0.1823

= 2.1942

Technology gives area = 0.4859

So,

P(19.9 < < 21.3) = 0.5 + 0.4859 = 0.9859

So,

Answer is:

0.9859

(e)

Correct option:

As sample size increased, the standard error decreased, resulting in higher probabilities.