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 20.6 hours. Assume these values have a normal distribution with a standard deviation of 1.3 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 21.1 hours?


(b) What is the probability that a randomly selected sample of 4 Rayovac batteries will have a mean test life between 19.9 and 21.1 hours?


(c) What is the probability that a randomly selected sample of 19 Rayovac batteries will have a mean test life between 19.9 and 21.1 hours?


(d) What is the probability that a randomly selected sample of 67 Rayovac batteries will have a mean test life between 19.9 and 21.1 hours?

Answer 1

This is a normal distribution question with
$\\Mean (\mu)= 20.6 \\Standard\;Deviation (\sigma)= 1.3 \\Since\; we\; know\; that \\z_{ score } = \frac{x-\mu}{\sigma}$
a) P(19.9 < x < 21.1)=?
$\\ The\; z-score\; at\; x = 19.9 is, \\ z = \frac{19.9-20.6}{1.3} \\ z_1 = -0.5385 \\ The\; z-score\; at\; x = 19.9 is, \\ z = \frac{21.1-20.6}{1.3} \\ z_2 = 0.3846$
This implies that
P(19.9 < x < 21.1) = P(-0.5385 < z < 0.3846) = P(Z < 0.3846) - P(Z < -0.5385)
P(19.9 < x < 21.1) = 0.64972 - 0.2951
P(19.9 < x < 21.1) = 0.3546

b) Sample size (n) = 4
Since we know that

Sample mean() 20.6 Sample standar d deviation (s) 0.65

P(19.9 < x < 21.1)=?
$\\ The\; z-score\; at\; x = 19.9 is, \\ z = \frac{19.9-20.6}{0.65} \\ z_1 = -1.0769 \\ The\; z-score\; at\; x = 19.9 is, \\ z = \frac{21.1-20.6}{0.65} \\ z_2 = 0.7692$
This implies that
P(19.9 < x < 21.1) = P(-1.0769 < z < 0.7692) = P(Z < 0.7692) - P(Z < -1.0769)
P(19.9 < x < 21.1) = 0.7791 - 0.1407
P(19.9 < x < 21.1) = 0.6384

c) Sample size (n) = 19
Since we know that
$\\Sample \; mean (\bar{x}) = \mu = 20.6 \\Sample \;standard\;deviation(s) = \frac{\sigma}{\sqrt{n}} = 0.2982$
P(19.9 < x < 21.1)=?
$\\ The\; z-score\; at\; x = 19.9 is, \\ z = \frac{19.9-20.6}{0.2982} \\ z_1 = -2.3474 \\ The\; z-score\; at\; x = 19.9 is, \\ z = \frac{21.1-20.6}{0.2982} \\ z_2 = 1.6767$
This implies that
P(19.9 < x < 21.1) = P(-2.3474 < z < 1.6767) = P(Z < 1.6767) - P(Z < -2.3474)
P(19.9 < x < 21.1) = 0.9532 - 0.0095
P(19.9 < x < 21.1) = 0.9437
d) Sample size (n) = 67
Since we know that
$\\Sample \; mean (\bar{x}) = \mu = 20.6 \\Sample \;standard\;deviation(s) = \frac{\sigma}{\sqrt{n}} = 0.1588$
P(19.9 < x < 21.1)=?
$\\ The\; z-score\; at\; x = 19.9 is, \\ z = \frac{19.9-20.6}{0.1588} \\ z_1 = -4.4081 \\ The\; z-score\; at\; x = 19.9 is, \\ z = \frac{21.1-20.6}{0.1588} \\ z_2 = 3.1486$
This implies that
P(19.9 < x < 21.1) = P(-4.4081 < z < 3.1486) = P(Z < 3.1486) - P(Z < -4.4081)
P(19.9 < x < 21.1) = 0.9992 - 5.2141e-06
P(19.9 < x < 21.1) = 0.9992
PS: you have to refer z score table to find the final probabilities.
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