No . . . while the difference represents the absolute magnitude between two numbers . . . for example . . .
<em>The difference between 5 and 2 is . . . 3</em>
<em>The difference between 6.4 and 9.5 is . . . 3.1</em>
. . . there is still the chance that the difference may be zero . . . in which case the difference is neither positive nor negative
. . . so in short . . . the answer is . . . <u><em>NO</em></u>
Answer:
7.3% of the bearings produced will not be acceptable
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
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:
Target value of .500 in. A bearing is acceptable if its diameter is within .004 in. of this target value.
So bearing larger than 0.504 in or smaller than 0.496 in are not acceptable.
Larger than 0.504
1 subtracted by the pvalue of Z when X = 0.504.
has a pvalue of 0.9938
1 - 0.9938= 0.0062
Smaller than 0.496
pvalue of Z when X = -1.5
has a pvalue of 0.0668
0.0668 + 0.0062 = 0.073
7.3% of the bearings produced will not be acceptable
Answer:
the anser would be the second one i think hope it helps please mark brailiest
Step-by-step explanation:
Answer: a) 1 / ⁴⁰C₅ b) 0.33
Step-by-step explanation:
a) The sample space consists of all numbers 1-40.
Since any of the number can be taken from the sample space so each of five 5 distinct numbers we take has equal probability of occurring. So probability of each 5 numbers set we take will be equal to 1 / ⁴⁰C₅
b)
If we pick exactly 3 even number then that means other 2 will be odd.
So, we have sample space of 40 numbers out of which 20 are even and 20 are odd.
Now we have to pick 3 even out of 20 and 2 odd out of 20.
Probability = ²⁰C₃ * ²⁰C₂ / ⁴⁰C₅
Probability= 0.33
