Question

Power+, produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal probability distribution with a mean of 27 hours and a standard deviation of 4.1 hours. As a part of its testing program, Power+ tests samples of 25 batteries. Use Appendix B.1 for the z-values. a. What can you say about the shape of the distribution of sample mean? Shape of the distribution is b. What is the standard error of the distribution of the sample mean? (Round the final answer to 4 decimal places.) Standard error c. What proportion of the samples will have a mean useful life of more than 28 hours? (Round the final answer to 4 decimal places.) Probability d. What proportion of the sample will have a mean useful life greater than 26.5 hours? (Round the final answer to 4 decimal places.) Probability e. What proportion of the sample will have a mean useful life between 26.5 and 28 hours? (Round the final answer to 4 decimal places.) Probability

Answer 1

a)

normal

b)

standard error = 4.1/sqrt(25) = 0.82

c)

Here, μ = 27, σ = 0.82 and x = 28. We need to compute P(X >= 28). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (28 - 27)/0.82 = 1.22

Therefore,
P(X >= 28) = P(z <= (28 - 27)/0.82)
= P(z >= 1.22)
= 1 - 0.8888 = 0.1112

d)

Here, μ = 27, σ = 0.82 and x = 26.5. We need to compute P(X >= 26.5). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (26.5 - 27)/0.82 = -0.61

Therefore,
P(X >= 26.5) = P(z <= (26.5 - 27)/0.82)
= P(z >= -0.61)
= 1 - 0.2709 = 0.7291

e)

Here, μ = 27, σ = 0.82, x1 = 26.5 and x2 = 28. We need to compute P(26.5<= X <= 28). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (26.5 - 27)/0.82 = -0.61
z2 = (28 - 27)/0.82 = 1.22

Therefore, we get
P(26.5 <= X <= 28) = P((28 - 27)/0.82) <= z <= (28 - 27)/0.82)
= P(-0.61 <= z <= 1.22) = P(z <= 1.22) - P(z <= -0.61)
= 0.8888 - 0.2709
= 0.6179