Shown below are the data giving the length of life of lights in an underground tunnel.

9.0 months 7.0 months
8.2 months 8.3 months
7.8 months 8.1 months
7.5 months 7.6 months
9.1 months 7.4 months

I calculated the standard deviation to be about 0.679.
Estimate the probability that a random chosen light will last as long as 7 months after installation. Assume the data to be described by a normal distribution.

Can anyone help me with the process I would use to calculate the answer? Thanks!

I assume that you have calculated the mean to find the SD.

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.

To estimate the probability that a randomly chosen light will last as long as 7 months after installation, you can use the concept of the standard normal distribution.

The first step is to calculate the z-score, which measures the number of standard deviations a particular data point is from the mean of the distribution. The formula for calculating the z-score is:

z = (x - μ) / σ

Where:
- z is the z-score
- x is the value you want to calculate the probability for (in this case, 7 months)
- μ is the mean of the distribution (in this case, the average lifespan of the lights)
- σ is the standard deviation of the distribution (0.679 months, as you calculated)

In this case, since the data is assumed to be normally distributed, you can use the mean and standard deviation calculated from the data as the parameters for the distribution.

Substituting the values:
z = (7 - μ) / 0.679

To calculate the probability, you will then look up the z-score on a standard normal distribution table or use a statistical calculator that provides this information. The resulting value will represent the probability of a randomly chosen light lasting as long as 7 months.

Keep in mind that the standard normal distribution assumes that the data is normally distributed, which might not always be the case in real-life situations.