1) A certain dimension of a mass-produced mechanical part has a nominal value of

Question

1) A certain dimension of a mass-produced mechanical part has a nominal value of 100
mm with acceptable tolerances of ± 1 mm. If the manufacturing process produces parts for which values of this dimension are normally distributed with mean 100.2 mm and standard deviation 0.5 mm, what percentage of parts will have to be rejected as outside the tolerance limits?

2) It has been found that, on weekdays, the speeds of free-moving vehicles on a certain
section of roadway are normally distributed with mean 72 km/h and standard
deviation 16 km/h. What is the speed that only 5% of the vehicles exceed?

3) The weight of active ingredient in 60-mg phenobarbitone tablets produced by a
pharmaceutical company is normally distributed with a mean of 58.5 mg and a
standard deviation of 1.2 mg. If a regulation states that each tablet should contain at
least 55.5 mg of active ingredient, what proportion of tablets fail to comply?

Explanation / Answer

Back-up Theory

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then,

Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(2)

Q1)

Let X = dimension of the mass-produced mechanical part. We are given

X ~ N(100.2, 0.52).

The dimension has a nominal value of 100 mm with acceptable tolerances of ± 1 mm. =>

LSL(Lower Specification Limit) = 100 – 1 = 99 and

USL(Lower Specification Limit) = 100 + 1 = 101.

Now, percentage percentage of parts will have to be rejected as outside the tolerance limits

= 100 x {P(X < 99) + P(X > 101)}

= 100 x {P[Z < {(99 – 100.2)/0.5}] + P[Z > {(101 – 100.2)/0.5}] [vide (2) of Back-up Theory]

= 100 x {P(Z < - 2.4) + P(Z > 1.6)}

= 100 x (0.0082 + 0.0548) [using Excel Function on Normal Distribution]

= 100 x 0.063

= 6.3% ANSWER

Q 2)

Let Y = the speeds of free-moving vehicles on a certain section of roadway. We are given Y ~ N(72, 162).

If t is the speed that only 5% of the vehicles exceed, then we should have

P(Y > t) = 0.05 or

P[Z > {(t – 72)/16}] = 0.05 [vide (2) of Back-up Theory]

=> {(t – 72)/16} = 1.645 [using Excel Function on Normal Distribution]

=> t = 72 + (1.645 x 16)

       = 72 + 26.32

       = 98.32

So, the speed that only 5% of the vehicles exceed is 98.32 ANSWER

Q 3)

Let X = the weight of active ingredient in 60-mg phenobarbitone tablets produced by the pharmaceutical company. We are given X ~ N(58.5, 1.22). Proportion of tablets that fail to comply with the regulation

= P(X < 55.5)

= P[Z < {(55.5 – 58.5)/1.2}] [vide (2) of Back-up Theory]

= P(Z < - 2.5)

= 0.0062 ANSWER [using Excel Function on Normal Distribution]

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.