Question

If a set of sample measurements has a mean of 100, a normal distribution, a standard deviation of 2, and control limits of 94 and 106, what percentage of the samples are expected to be between 94 and 106? Explain your answer.

Answer 1

Answer: 99.73%

Step-by-step explanation:

Given : Mean : $\mu=100$

Standard deviation : $\sigma=2$

Let X be the random variable that represents the data values.

Formula for Z-score :  $z=\dfrac{X-\mu}{\sigma}$

For x=94, we have

$z=\dfrac{94-100}{2}=-3$

For x=106, we have

$z=\dfrac{106-100}{2}=3$

The probability that the samples are between 94 and 106:-

$P(-3$

Hence, the percent of the samples are expected to be between 94 and 106 = 99.73%