Answer:
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)
Answer:
1.22
Step-by-step explanation:
Since there is one whole shaded. And 2 tens, and 2 ones. Its 1.22.
Answer:
The answer is -51146
Step-by-step explanation:
Just substract simple
Answer:
All rhombuses are rectangles. False
All parallelograms are quadrilaterals. True
All squares are rhombuses. True
All rectangles are squares. False
Answer:
A
Step-by-step explanation:
