Question

A particular type of ballpoint pen uses minute ball bearings that are targeted to
have a diameter of 0.5 mm. The lower and upper speci cation limits under which the
ball bearing can operate are 0.49 mm (lower) and 0.51 mm (upper). Past experience
has indicated that the ball bearings are approximately normally distributed with a
mean of 0.503 mm and a standard deviation of 0.004 mm. If you select a sample of 25
ball bearings, what is the probability that the sample mean is:
(a) between the target and the population mean of 0.503 mm?
(b) between the lower speci cation limit and the target?
(c) above the upper speci cation?
(d) The probability is 93.32% that the sample mean diameter will be above what value?

Answer 1

Target diametr = 0.5 mm

Lower specification = 0.49 mm

Upper specification = 0.51 mm

Here te ball bearings are normally distibutioe with mean = 0.503 mm and standard deviaion = 0.004 mm.

Sample size = 25

standard error of sample mean = 0.004/sqrt(25) = 0.0008 mm

(a) If is the sample mean of a random sample of size 25 than

Pr(0.5 mm < < 0.503 mm) = Pr( < 0.503 mm ; 0.503 mm ; 0.0008 mm) - Pr( < 0.50 mm; 0.503 mm ; 0.0008 mm)

Z₂ = 0

Z₁ = (0.50 - 0.503)/0.0008 = -3.75

Pr(0.5 mm < < 0.503 mm) = Pr( < 0.503 mm ; 0.503 mm ; 0.0008 mm) - Pr( < 0.50 mm; 0.503 mm ; 0.0008 mm) = Pr(Z < 0) - Pr(Z < -3.75) = 0.5 - 0.00009 = 0.49991

(b) Pr(0.49 mm < < 0.50 mm) = Pr( < 0..50 mm 0.503 mm ; 0.0008 mm) - Pr( < 0..49 mm; 0.503 mm ; 0.0008 mm)

Z₂ = -3.75

Z₁ = (0.49 - 0.503)/0.0008 = -16.25

Pr(0.49 mm < < 0.50 mm) = Pr(Z < -3.75) - Pr(Z < -16.25) = 0.00009 - 0 = 0.00009

(c) Pr( > 0.51 mm) = 1 - Pr( < 0.51 mm ; 0.503 mm ;0.0008 mm)

Z = (0.51 - 0.503)/0.0008 = 8.75

Pr( > 0.51 mm) = 1 - Pr( < 0.51 mm ; 0.503 mm ;0.0008 mm) = 1 - Pr(Z < 8.75) = 1 - 1 ~ 0

(d) Here let say that value is

Pr( > ) = 1- Pr( < ; 0.503 mm ; 0.0008 mm) = 0.9332

Pr( < ; 0.503 mm ; 0.0008 mm) = 0.0668

Here the Z value for the given p- value

Z = -1.5

-1.5 = ( - 0.503)/0.0008

= 0.503 - 1.5 * 0.0008 = 0.5018 mm