Question

The lifetimes of a certain type of car battery are normally distributed with a mean 5.9 years and standard deviation 0.4 year. The batteries are guaranteed to last at least 5 years. What proportion of batteries fail to meet the guarantee? In other words, what proportion of batteries last less than 5 years?

Answer 1

Answer:

1.22% of batteries last less than 5 years

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 5.9, \sigma = 0.4$

What proportion of batteries fail to meet the guarantee? In other words, what proportion of batteries last less than 5 years?

This is the pvalue of Z when X = 5. So

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

$Z = \frac{5 - 5.9}{0.4}$

$Z = -2.25$

$Z = -2.25$ has a pvalue of 0.0122

1.22% of batteries last less than 5 years

Answer 2

Answer:

$P(X$

And we can find this probability using the normal standard table or excel and we got:

$P(z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the lifetimes of a population, and for this case we know the distribution for X is given by:

$X \sim N(5.9,0.4)$  

Where $\mu=5.9$ and $\sigma=0.4$

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using the normal standard table or excel and we got:

$P(z$