The diameter of Ping-Pong balls manufactured at a large factory is expected to b

Question

The diameter of Ping-Pong balls manufactured at a large factory is expected to be approximately normally distributed with a mean of 1.30 inches and a standard deviation of .04 inch. If many random samples of 16 Ping-Pong balls are selected.

a) at what diameter will no more than 60% of the sample means be?

b) which is more likely to occur, an individual ball above 1.34 inches, a sample mean above 1.32 inches in a sample size of 4, or a sample mean above 1.31 inches in a sample size of 16? Explain

Explanation / Answer

Solution

Let X = diameter of Ping-pong ball. We are given, X has Normal Distribution.

All probabilities are evaluated using Excel Function.

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)

X bar ~ N(µ, 2/n),…………………………………………………………….…….(3),

where X bar is average of a sample of size n from population of X.

So, P(X bar or t) = P[Z or {(n)(t - µ)/ }] …………………………………(4)

Now, to work out solution,

Given µ = 1.3 and = 0.04

Part (a)

Given n = 16, [vide (3) under Back-up Theory], Xbar ~ N(1.3, 0.042/16).

Diameter at which no more than 60% of the sample means will be = d, say. Then, we should have: P(Xbar < d) 0.6 => [vide (4) under Back-up Theory], P[Z < {(d – 1.13)/0.01}] 0.6

=> {(d – 1.3)/0.01} = 0.2533 => d = 1.3025 ANSWER

Part (b)

P(X > 1.34) = P[Z > {(1.34 – 1.3)/0.04} = P(Z > 1) = 0.1587 ANSWER 1

Given n = 4, P(Xbar > 1.32) = P[Z > {(1.32 – 1.3)/0.02} = P(Z > 1) = 0.1587 ANSWER 2

Given n = 16, P(Xbar > 1.31) = P[Z > {(1.31 – 1.3)/0.01} = P(Z > 1) = 0.1587 ANSWER 3

So, all three are equally likely. ANSWER

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