Engr. 323
Prob. 5.81
Benjamin Dolf
La Devoir Belle Numero 14
5.81) Let µ denote the true pH of a chemical compound. A sequence of n independent sample pH
determinations will be made. Suppose each sample pH is a random variable with expected
value µ and standard deviation 0.1. How many determinations are required if we wish the
probablility that the sample average is within .02 of the true pH to be at least .95? What
theorem justifies your probablility calculation?
Let’s start with the Central Limit Theorem as shown on page 232 of the text The CLT states that if we take
n samples XI from a population with a mean of µ and a standard deviation of σ and let X = X1 + X2 +
X3...+ Xn , Then we can say
X ~ N( µ, σ
). For n > 30.
n
Solution:
Using the CLT we know E( X ) = µ and V( X ) = σ2/n =.12/n
Let X ~ N ( µ = µ, σ 2 = σ2/n =.12/n)
X ~ N(µ , σ= .1
n
)
This is a percentile problem, where we will use the Z tables to find a value for n. The question asks us to
find a value for n so that the distribution is within .02 of the mean. Using techniques from chapter 4.3:


(− .02 + µ ) − µ
(.02 + µ ) − µ 
0.95 = P(µ - .02 ≤ X ≤µ + .02) = P
≤Z .025 ≤
= .95
.1
.1


n
n


(
)
canceling the µ’s: P − 0.2 n ≤ Z .025 ≤0.2 n = .95
The Z values where 95% of the graph lie are –1.96, and 1.96.
P (− 1.96 ≤ Z ≤1.96 )= .95
You can find them by looking up Z.025 = -1.96 or Z.975 = 1.96.
Then
0.2 n = 1.96
Solving for n, we find that the required number of samples is n = 97.
1
Engr. 323
Prob. 5.81
Benjamin Dolf
La Devoir Belle Numero 14
As mentioned above, the Central Limit Theorem justifies our answer.
Normal Distribution of Z
0.45
0.4
-1.96
0.35
1.96
f(Z)
0.3
0.25
f(Z)
0.2
0.15
0.1
Z(.025)
0.05
Z(.975)
0
-6
-4
-2
0
2
4
Z
The area of interest lies bewteen the two lines
2
6