Question

In a study of plant safety, it was found that the time it took for machine operators to react to a warning light was normally distributed with a mean 1 second and standard deviation 0.3 second. Suppose a warning light visible to 4 operators goes on. What is the probability that the average (mean) reaction time of the 4 operators exceeds 1.25 seconds?

Answer 1

Answer:

4.75% probability that the average (mean) reaction time of the 4 operators exceeds 1.25 seconds.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Normal probability distribution

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:

$\mu = 1, \sigma = 0.3, n = 4, s = \frac{0.3}{\sqrt{4}} = 0.15$

What is the probability that the average (mean) reaction time of the 4 operators exceeds 1.25 seconds?

This is 1 subtracted by the pvalue of Z when X = 1.25. So

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

$Z = \frac{1.25 - 1}{0.15}$

$Z = 1.67$

$Z = 1.67$ has a pvalue of 0.9525.

So there is a 1-0.9525 = 0.0475 = 4.75% probability that the average (mean) reaction time of the 4 operators exceeds 1.25 seconds.