Question

A certain type of automobile battery is known to last an average of 1,150 days with a standard deviation of 40 days. If 100 of these batteries are selected, find the following probabilities for the average length of life of the selected batteries. (Round your answers to four decimal places.) A button hyperlink to the SALT program that reads: Use SALT. (a) The average is between 1,142 and 1,150. (b) The average is greater than 1,158. (c) The average is less than 950.

Answer 1

Answer:

a) 0.4772 = 47.72% probability that the average is between 1,142 and 1,150.

b) 0.0228 = 2.28% probability that the average is greater than 1,158.

c) 0 = 0% probability that the average is less than 950.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

A certain type of automobile battery is known to last an average of 1,150 days with a standard deviation of 40 days.

This means that $\mu = 1150, \sigma = 40$

Sample of 100:

This means that $n = 100, s = \frac{40}{\sqrt{100}} = 4$

(a) The average is between 1,142 and 1,150.

This is the pvalue of Z when X = 1150 subtracted by the pvalue of Z when X = 1142. So

X = 1150

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{1150 - 1150}{4}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

X = 1142

$Z = \frac{X - \mu}{s}$

$Z = \frac{1142 - 1150}{4}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228

0.5 - 0.0228 = 0.4772

0.4772 = 47.72% probability that the average is between 1,142 and 1,150.

(b) The average is greater than 1,158.

This is 1 subtracted by the pvalue of Z when X = 1158. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{1158 - 1150}{4}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

1 - 0.9772 = 0.0228

0.0228 = 2.28% probability that the average is greater than 1,158.

(c) The average is less than 950.

This is the pvalue of Z when X = 950. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{950 - 1150}{4}$

$Z = -50$

$Z = -50$ has a pvalue of 0

0 = 0% probability that the average is less than 950.