The lifetime of a particular type of battery is normally distributed with a mean of 1100 days
and a standard deviation of 80 days. The manufacturer randomly selects 400 batteries of this
type and ships them to a departmental store.
(a) What is the mean and standard deviation of the sampling distribution of X ?
(b) What is the probability that the average lifetime of these 400 batteries is between 1097
and 1104 days ?
To answer these questions, we will use the Central Limit Theorem (CLT).
(a) The mean of the sampling distribution of X is equal to the mean of the population, which is 1100 days. The standard deviation of the sampling distribution of X is given by the formula:
Standard Deviation of Sampling Distribution = Standard Deviation of Population / sqrt(sample size)
In this case, the standard deviation of the sampling distribution is 80 / sqrt(400) = 4.
Therefore, the mean of the sampling distribution of X is 1100 days and the standard deviation is 4 days.
(b) To find the probability that the average lifetime of these 400 batteries is between 1097 and 1104 days, we need to standardize these values using the sampling distribution.
Z = (X - Mean of Sampling Distribution) / Standard Deviation of Sampling Distribution
For 1097 days:
Z1 = (1097 - 1100) / 4 = -0.75
For 1104 days:
Z2 = (1104 - 1100) / 4 = 1
Now, we need to find the probability that Z is between -0.75 and 1. Using a standard normal distribution table or a calculator, we can look up the probabilities associated with these Z-values.
P(-0.75 ≤ Z ≤ 1) = P(Z ≤ 1) - P(Z ≤ -0.75)
From the Z-table, we find:
P(Z ≤ 1) = 0.8413
P(Z ≤ -0.75) = 0.2266
Therefore, P(-0.75 ≤ Z ≤ 1) = 0.8413 - 0.2266 = 0.6147
So, the probability that the average lifetime of these 400 batteries is between 1097 and 1104 days is 0.6147 (or 61.47%).
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