The diameters of ball bearings are distributed normally. The mean diameter is 87 millimeters and the standard deviation is 6 millimeters. Find the probability that the diameter of a selected bearing is greater than 84 millimeters. Round your answer to four decimal places.

69.14% probability that the diameter of a selected bearing is greater than 84 millimeters

Step-by-step explanation:

According to the Question,

Given That, The diameters of ball bearings are distributed normally. The mean diameter is 87 millimeters and the standard deviation is 6 millimeters. Find the probability that the diameter of a selected bearing is greater than 84 millimeters.

In a set with mean and standard deviation, the Z score of a measure X is given by Z = (X-μ)/σ

we have μ=87 , σ=6 & X=84

Find the probability that the diameter of a selected bearing is greater than 84 millimeters

This is 1 subtracted by the p-value of Z when X = 84.

So, Z = (84-87)/6

Z = -3/6

Z = -0.5 has a p-value of 0.30854.

⇒1 - 0.30854 = 0.69146

0.69146 = 69.14% probability that the diameter of a selected bearing is greater than 84 millimeters.

Note- (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 p-value, we get the probability that the value of the measure is greater than X)

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Num of plants = (original#ofplants)*(1.6)^t