Answer:
(a) The proportion of the diameters are less than 25.0 mm is 0.1056.
(b) The 10th percentile of the diameters is 24.99 mm.
(c) The ball bearing that has a diameter of 25.2 mm is at the 84th percentile.
(d) The proportion of the ball bearings meeting the specification is 0.8881.
Step-by-step explanation:
Let <em>X</em> = diameters of ball bearings.
The random variable <em>X</em> is normally distributed with mean, <em>μ</em> = 25.1 mm and standard deviation, <em>σ</em> = 0.08 mm.
To compute the probability of a Normally distributed random variable we need to first convert the raw scores to <em>z</em>-scores as follows:
<em>z</em> = (X - μ) ÷ σ
(a)
Compute the probability of <em>X</em> < 25.0 mm as follows:
P (X < 25.0) = P ((X - μ)/σ < (25.0-25.1)/0.08)
= P (Z < -1.25)
= 1 - P (Z < 1.25)
= 1 - 0.8944
= 0.1056
*Use a <em>z</em>-table for the probability.
Thus, the proportion of the diameters are less than 25.0 mm is 0.1056.
(b)
The 10th percentile implies that, P (X < x) = 0.10.
Compute the 10th percentile of the diameters as follows:
P (X < x) = 0.10
P ((X - μ)/σ < (x-25.1)/0.08) = 0.10
P (Z < z) = 0.10
<em>z</em> = -1.282
The value of <em>x</em> is:
z = (x - 25.1)/0.08
-1.282 = (x - 25.1)/0.08
x = 25.1 - (1.282 × 0.08)
= 24.99744
≈ 24.99
Thus, the 10th percentile of the diameters is 24.99 mm.
(c)
Compute the value of P (X < 25.2) as follows:
P (X < 25.2) = P ((X - μ)/σ < (25.2-25.1)/0.08)
= P (Z < 1.25)
= 0.8944
≈ 0.84
*Use a <em>z</em>-table for the probability.
Thus, the ball bearing that has a diameter of 25.2 mm is at the 84th percentile.
(d)
Compute the value of P (25.0 < X < 25.3) as follows:
P (25.0 < X < 25.3) = P ((25.0-25.1)/0.08 < (X - μ)/σ < (25.3-25.1)/0.08)
= P (-1.25 < Z < 2.50)
= P (Z < 2.50) - P (Z < -1.25)
= 0.99379 - 0.10565
= 0.88814
≈ 0.8881
*Use a <em>z</em>-table for the probability.
Thus, the proportion of the ball bearings meeting the specification is 0.8881.
