Answer:
7.3%
Step-by-step explanation:
to get this percentage, we need to use the z-score calculation;
z-score = (x-mean)/SD
mean = 25.10 , SD = 0.2
For diameter 24.60
z-score = (24.6-25.10)/0.2 = -0.5/0.2 = -2.5
For diameter 25.40
z-score = (25.4-25.1)/0.2 = 1.5
So the proportion that will satisfy the specifications will be;
P(-2.5 < z < 1.5)
At this point, we use the standard normal table
P(-2.5 < z < 1.5) = P (z<1.5) - P (Z < -2.5)
From standard normal table;
P(Z < 1.5) = 0.9332
P(z < -2.5) = 0.0062
P(-2.5 < z < 1.5) = 0.9332 - 0.0062 = 0.927
So the proportion that meets specification = 92.7% ( same as 0.927)
Proportion failing to meet specification = 100- 92.7% = 7.3%
Answer: The percentage of ball bearings fail to satisfy the contract specifaction is 7.3 %.
Step-by-step explanation:
First we need to find the standard score of 24.60 and 25.40. Using Standard Score formula.
[tex]\frac{24.60 - 25.10}{.20} =-2.5\\\\\frac{25.4 - 25.10}{.20}=1.5[/tex]
now we need to find the percentages of these Z scores.
-2.5 = .0062 or 00.62%
1.5 = .9332 or 93.32%
now we subtract these two to get the in between percentage.
.9332 - .0062 = .927 or 92.7% is what doesnt fail to satifsy the contract spefications! But to find what does fail to satisfy we have to subtract it by 100% or 1.
1 - .927 = 0.073
7.3%
