Question

The lifetime of a certain battery is normally distributed with a mean value of 20 hours and a standard deviation of 2.5 hours.

a. What are the distribution parameters (μ and σ) of the sample mean if you sample a four pack of batteries from this population?

b. If there are four batteries in a pack, what is the probability that the average lifetime of these four batteries lies between 18 and 20?

c. What happens to the probability in part 1.b. if the number of batteries in the sample goes up?

Show all work please

Answer 1

a)
mean = 20

std.deviation = 2.5/sqrt(4) = 1.25

b)
Here, μ = 20, σ = 1.25, x1 = 18 and x2 = 20. We need to compute P(18<= X <= 20). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (18 - 20)/1.25 = -1.6
z2 = (20 - 20)/1.25 = 0

Therefore, we get
P(18 <= X <= 20) = P((20 - 20)/1.25) <= z <= (20 - 20)/1.25)
= P(-1.6 <= z <= 0) = P(z <= 0) - P(z <= -1.6)
= 0.5 - 0.0548
= 0.4452

c)

Here, μ = 20, σ = 0.625, x1 = 18 and x2 = 20. We need to compute P(18<= X <= 20). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (18 - 20)/0.625 = -3.2
z2 = (20 - 20)/0.625 = 0

Therefore, we get
P(18 <= X <= 20) = P((20 - 20)/0.625) <= z <= (20 - 20)/0.625)
= P(-3.2 <= z <= 0) = P(z <= 0) - P(z <= -3.2)
= 0.5 - 0.0007
= 0.4993

if we increase the sample size th eprobability will increase Here, we tae sample size = 16