Question

A standard fluorescent tube has a life length that is normally distributed with a mean of 7000 hours and a standard deviation of 1000 hours. A competitor has developed a compact fluorescent lighting system that will fit into incandescent sockets. It claims that a new compact tube has a normally distributed life length with a mean of 7500 hours and a standard deviation of 1200 hours.
1. Which fluorescent tube is more likely to have a life length greater than 9000 hours?

Answer 1

Answer:

The new compact tube is more likely to have a life length greater than 9000 hours.

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:

Standard tube:

$\mu = 7000, \sigma = 2000$

New compact tube:

$\mu = 7500, \sigma = 1200$

1. Which fluorescent tube is more likely to have a life length greater than 9000 hours?

Whichever tube has the lower z-score when X = 9000, since the probability of having a life length greater than 9000 hours is 1 subtracted by the pvalue of Z when X = 9000. The higer z, the higher it's pvalue.

Standard tube:

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

$Z = \frac{9000 - 7000}{1000}$

$Z = 2$

New compact tube:

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

$Z = \frac{9000 - 7500}{1200}$

$Z = 1.25$

The new compact tube is more likely to have a life length greater than 9000 hours.