Question

A manufacturer claims that a certain brand of VCR player has an average life expectancy of 5 years and 6 months with a standard deviation of 1 year and 6 months. Assume that the life expectancy is normally distributed.14. Based on the information above, when selecting one VCR player at random what is the probability of its life expectancy being greater than seven years

Answer 1

Answer:

15.87% probability of its life expectancy being greater than seven years

Step-by-step explanation:

Problems of normally distributed samples are 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:

Life expectancy of 5 years and 6 months

6 months is half an year, so $\mu = 5.5$

Standard deviation of 1 year and 6 months.

So $\sigma = 1.5$

Probability of its life expectancy being greater than seven years

This is 1 subtracted by the pvalue of Z when X = 7.

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

$Z = \frac{7 - 5.5}{1.5}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

1 - 0.8413 = 0.1587

15.87% probability of its life expectancy being greater than seven years