Download Supplementary Exercises Sampling Distribution

Supplementary Exercises
Sampling Distribution of the Sample Mean
1. A bottling company uses a filling machine to fill plastic bottles with a popular cola. The
bottles are supposed to contain 300 milliliters. In fact, the contents vary according to a
normal distribution with mean µ = 298 ml and standard deviation ) = 3 ml.
a. What is the probability that an individual bottle contains less than 295 ml?
b. What is the probability that the mean contents of the bottles in a six-pack is less than 295
ml?
2. The level of nitrogen oxide (NOX) in the exhaust of a particular car model varies with mean
1.4 g/mile and standard deviation 0.3 g/mile. A company has 125 cars of this model in its
fleet. If X̄ is the mean NOX emission level for these 125 cars, what is the level L such that
the probability that X̄ is greater than L is only 0.01?
3. The number of flaws per square yard in a type of carpet material varies, with mean 1.6 flaws
per square yard and standard deviation 1.2 flaws per square yard. The distribution is not
normal--in fact, it is discrete. An inspector studies 200 random square yards of the material,
records the number of flaws found in each square yard, and calculates the mean number of
flaws per square yard inspected. Use the central limit theorem to find the approximate
probability that the mean number of flaws exceeds 2 per square yard.
4. The number of accidents per week at a hazardous intersection varies with mean 2.2 and
standard deviation 1.4. This distribution is discrete and so is certainly not normal.
a. Let X̄ be the mean number of accidents per week at the intersection during a year (52
weeks). According to the central limit theorem, what is the approximate distribution of
X̄ ?
b. What is the approximate probability that there are fewer than 100 accidents at the
intersection in a year? (HINT: Restate this event in terms of X̄ .)
/ Solutions on next page.
Solutions
1a. P(X < 295) = P[Z < (295298)/3] = P(Z < -1.00) = 0.1587
1b. )x̄ )
n
3
1.225
6
So, P( X̄ < 295) = P[Z < (295298)/1.225] = P(Z < 2.45) = 0.0071
2. Find L such that P( X̄ > L) = 0.01.
This is equivalent to finding L such that P( X̄ < L) = 1 0.01= 0.99.
And, this is equivalent to finding zo such that P(Z < zo ) = 1 0.01= 0.99. Using Table III,
the normal distribution table, we find that zo = 2.33.
Now, we just need to transform this back to X̄ .
X̄ µ
)
Since Z , we can rearrange this so that X̄ µ zo
.
)/ n
n
That is, X̄ = 1.4 + 2.33(0.3/11.18) = 1.463.
3. Since the sample size is larger than 30 (n = 200), the Central Limit Theorem applies. So,
)
1.2
)x̄ 0.085
n
200
So, P( X̄ > 2) = P[Z > (21.6)/0.085] = P(Z > 4.71) ³ 0.
4a. Since the sample size is larger than 30 (n=52), the Central Limit Theorem applies. So, X̄ is
)
1.4
approximately normally distributed with mean µ = 2.2 and )x̄ 0.194 .
n
52
4b. P(total number of accidents < 100) = P(* x < 100) = P( X̄ < 100/52) = P( X̄ < 1.923)
= P(Z < (1.9232.2)/0.194) = P(Z < 1.43) = 0.0764