For a standard normally distributed random variable <em>Z</em> (with mean 0 and standard deviation 1), we get a probability of 0.0625 for a <em>z</em>-score of <em>Z</em> ≈ 1.53, since
P(<em>Z</em> ≥ 1.53) ≈ 0.9375
You can transform any normally distributed variable <em>Y</em> to <em>Z</em> using the relation
<em>Z</em> = (<em>Y</em> - <em>µ</em>) / <em>σ</em>
where <em>µ</em> and <em>σ</em> are the mean and standard deviation of <em>Y</em>, respectively.
So if <em>s</em> is the standard deviation of <em>X</em>, then
(250 - 234) / <em>s</em> ≈ 1.53
Solve for <em>s</em> :
16/<em>s</em> ≈ 1.53
<em>s</em> ≈ 10.43
2 out of 50 are defective.
Divide 2 by 50: 2/50 = 0.04
Now multiply the total quantity by that:
2000 x 0.04 = 80
80 are likely to be defective.
Answer:
12.568m^2
Step-by-step explanation:
A=πr^2 where r is. radius
=3.142(2^2)
=3.142(4)
=12.568m^2
A) x/y = 2/3
B) x + y = 105
Solving Equation A for y
A) y = 1.5x
Substituting A into B
B) x + 1.5x = 105
2.5x = 105
x = 42
y = 63
