Question

In an inventory of standardized weights, 8% of the weights are found to exceed 10.282 grams and 6% weighs below 9.69 grams. If weight is a continuous random variable distributed normally, find the mean and standard deviation for this distribution.

Answer 1

Explanation:

We are given here for the random variable X:

P( X > 10.282) = 0.08 and P(X < 9.69) = 0.06

From standard normal tables, we have:

P(Z < 1.405) = 0.92. Therefore P(Z > 1.405) = 0.08

Therefore, the z score of 10.282 is 1.405

Therefore, Mean + 1.405*Std Dev = 10.282

Also from standard normal tables, we have:

P(Z < -1.555) = 0.06

Therefore, Mean - 1.555×Std Dev = 9.69

Subtracting the second equation from first, we get here:

Std Dev(1.405 + 1.555) = 10.282 - 9.69

Std Dev = 0.2

Now the mean can be computed as:

Mean = 1.555*Std Dev + 9.69 = 1.555*0.2 + 9.69 = 10.001

Therefore 10.001 is the required mean and 0.2 is the required standard deviation for the distribution here.