Download Name March 8, 2007 Math 145 - Elementary Statistics Long Exam II

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Math 145 - Elementary Statistics
March 8, 2007
Long Exam II
Instructions: Please include your work to get full credit and write your solutions neatly. Make sure that
you round off the decimal values correctly. Encircle your final answers.
1. Let X be the number of cars that Linda sells on a typical Saturday. The probability distribution for
X is given below
Car sold (x)
Probability
0
0.15
1
0.4
2
0.25
3
0.15
4
0.05
a. What is the probability that on a typical Saturday, Linda will sell at least 2 cars?
[3]
P (X ≥ 2) = 0.25 + 0.15 + 0.05 = 0.45
b. Find the mean (µX ) and standard deviation (σX ) of X.
[6]
µX = E(X) = 0(.15) + 1(.4) + 2(.25) + 3(.15) + 4(.05) = 1.55
2
σX
= V (X) = (0 − 1.55)2 (.15) + (1 − 1.55)2 (.4) + · · · + (4 − 1.55)2 (.05)
σX
= .360 + .121 + .051 + .315 + .300 = 1.147
√
1.147 ≈ 1.07.
=
c. If Linda is paid $50 per day and earns a commission of $250 for each car she sells, what is her
expected earning on a typical Saturday?
[3]
E(Earning)=E(250X + 50) = 250E(X) + 50 = 250(1.55) + 50 ≈ $437.50
d. If Linda is given a choice to earn $100 per day but with a lower commission of $200 for each car
she sells, determine which choice (compared to that of part c) is better in the long run.
[4]
E(Earning)=E(200X + 100) = 200E(X) + 100 = 200(1.55) + 100 = $410
Hence, the choice with the higher commission and $50 fixed salary is better in the long run.
2. A company that produces fine crystal knows from experience that 10% of its goblets have cosmetic
flaws and must be classified as “seconds.” Among six randomly selected goblets, what is the probability that
a. not one is a second?
Let X = no. of “seconds” out of 6. Then X ∼ Binomial(n = 6, p = .10).
6
P (X = 0) =
(.10)0 (.90)6 ≈ 0.5314.
0
[4]
b. at least three are seconds?
[4]
P (X ≥ 3) = 1 − P (X ≤ 2) = 1 − 0.984 = 0.016.
3. In response to concerns about nutritional contents of fast foods, McDonald’s has announced that it
will use a new cooking oil for its french fires that will decrease substantially trans fatty acid levels
and increase the amount of more beneficial poly-unsaturated fat. The company claims that 97 out
of 100 people cannot detect a difference in taste between the new and old oils. Assuming that this
figure is correct (as a long-run proportion), and let X denote the number of people who can tell the
difference between the two oils out of a random sample of 1000 individuals who purchased fries at
McDonald’s,
a. Determine the exact distribution of X. Specify all known distribution parameter values.
[3]
X ∼ Binomial(n = 1000, p = 0.03).
b. Explain why the normal distribution can be used to approximate this distribution, then give
parameter values of the approximate normal distribution.
[3]
√
Since np = 30 and n(1 − p) = 970, both are > 10, then X ≈ N (µ = 30, σ = 29.1).
c. Approximate the probability that at least 40 people (out of the 1000) can tell the difference
between the two oils.
[4]
X −µ
39.5 − 30
P (X ≥ 40) ≈ P
≥ √
σ
29.1
= P (Z > 1.76)
= 1 − .9608 = 0.0392.
d. Approximate the probability that between 20 to 34 (inclusive) can tell the difference between
the two oils.
[5]
X −µ
34.5 − 30
19.5 − 30
√
≤
≤ √
P (20 ≤ X ≤ 34) ≈ P
σ
29.1
29.1
= P (−1.95 < Z < .83)
= .7967 − .0256 = 0.7711.
4. The average hourly wage of fast-food workers employed by a nationwide chain is $6.75 and the
standard deviation is $1.15. Assume that the hourly wage is normally distributed.
a. Then about 68% of all such workers earn between 6.75−1.15=$5.6 and 6.75+1.15=$7.9 dollars
per hour.
[2]
b. What proportion of these workers make at least $7.00 per hour?
7 − 6.75
X −µ
≥
P (X ≥ 7) = P
σ
1.15
= P (Z > 0.22)
[4]
= 1 − .5871 = 0.4129.
c. What proportion of these workers earn between $6.00 to $7.00 per hour?
6 − 6.75
X −µ
7 − 6.75
P (6 ≤ X ≤ 7) = P
≤
≤
1.15
σ
1.15
= P (−0.65 < Z < .22)
[5]
= .5871 − .2578 = 0.3293.
d. Under these conditions, how much should your hourly wage be in order to be in the top 10%
among such workers?
[4]
Let x10% be the minimum hourly wage that belongs to the top 10%. That is,
X −µ
x10% − 6.75
P (X ≥ x10% ) = 0.10 =⇒ P
≥
= 0.10
σ
1.15
x10% − 6.75
x10% − 6.75
=⇒
P Z≥
= 0.10 =⇒ P Z <
= 0.90.
1.15
1.15
x
− 6.75
But
P (Z < 1.28) ≈ 0.90. Hence, 10%
= 1.28
1.15
T heref ore, x10% = 1.28(1.15) + 6.75 = 8.22.
e. If 25 of these workers are randomly selected and let X̄ represent the average salary of these 25
workers, what is the exact distribution of X̄? Specify all known parameters.
[3]
1.15
X̄ ∼ N µ = 6.75, σX̄ = √
.
25
f. What is the probability that the average hourly wage of those 25 workers is
i. at least $7.00 per hour.
[4]
7 − 6.75
X̄ − µ
√
√ ≥
σ/ n
1.15/ 25
= P (Z > 1.09)
P (X̄ ≥ 7) = P
= 1 − .8621 = 0.1379.
ii. between $6.00 to $7.00 per hour?
[5]
6 − 6.75
X̄ − µ
7 − 6.75
√ ≤
√
√ ≤
P (6 ≤ X̄ ≤ 7) = P
σ/ n
1.15/ 25
1.15/ 25
= P (−3.26 < Z < 1.09)
= .8621 − .0006 = 0.8615.
5. There are two machines (M1 and M2 ) available for cutting corks intended for use in wine bottles.
Machine M1 produces corks with diameters that are normally distributed with mean 3 cm and standard deviation 0.06 cm. Machine M2 produces corks with diameters that have a normal distribution
with mean 3.04 cm and a standard deviation 0.02 cm.
a. If a cork is randomly selected from M1 , what is the probability that the diameter of this cork is
longer than 3.05 cm?
[4]
M1 : X ∼ N (µX = 3, σX = 0.06)
X − µX
3.05 − 3
P (X ≥ 3.05) = P
≥
≈ P (Z > .83)
σX
0.06
= 1 − .7967 = 0.2033.
b. If a cork is randomly selected from M2 , what is the probability that the diameter of this cork is
longer than 3.05 cm?
[4]
∼ N (µY = 3.04, σY = 0.02)
Y − µY
3.05 − 3.04
P (X ≥ 3.05) = P
≥
≈ P (Z > .5)
σY
0.02
= 1 − .6915 = 0.3085.
M2 : Y
c. If acceptable corks have diameters between 2.9 cm and 3.1 cm, compute the proportions of acceptable corks produced by M1 and M2 . Which machine is more likely to produce an acceptable
cork?
[7]
2.9 − 3
X − µX
3.1 − 3
M1 : P (2.9 < X < 3.1) = P
<
<
≈ P (−1.67 < Z < 1.67)
0.06
σX
0.06
= 0.9525
− 0.0475 = 0.905.
2.9 − 3.04
Y − µY
3.1 − 3.04
M2 : P (2.9 < Y < 3.1) = P
<
<
= P (−7 < Z < 3)
0.02
σY
0.02
= 0.9987 − 0 = 0.9987.
Therefore, Machine 2 is more likely to produce acceptable cork.
d. If 16 corks from M1 are randomly selected, what is the probability the the average diameter of
these 16 corks is bigger than 3.05 cm?
[6]
0.06
X̄ ∼ N µX̄ = 3, σX̄ = √
16
X̄ − µX̄
3.05 − 3
√ ≥
√
P (X̄ ≥ 3.05) = P
≈ P (Z > 3.33)
σX / n
0.06/ 16
= 1 − .9996 = 0.0004.
6. The scores of high school seniors on the ACT college entrance examination in 2003 had a mean of
20.6 and standard deviation of 4.8. If a random sample of 36 students is selected
a. What can you say about the distribution of the average of the scores of these 36 students?
What result guarantees the veracity of your statement? Don’t forget to include the appropriate
parameter values
[4]
4.8
√
X̄ ≈ N µ = 20.6, σX̄ =
36
Since n is large (> 30), the approximate normality of X̄ is guaranteed by the Central Limit
Theorem.
b. What is the probability that the average score of the 36 students is lower than 18?
X̄ − µ
18 − 20.6
√ <
√
P (X̄ < 18) = P
σ/ n
4.8/ 36
= P (Z < −3.25) = .0006.
7. Answer the following questions with at most 3 sentences.
[6]
[3 pts. each]
a. Explain the difference between a parameter and a statistic? Give 2 examples of each.
A parameter is a numerical descriptive measure of a population and is constant at a given time.
Examples of parameters are µ and σ 2 . On the other hand, a statistic is a numerical descriptive
measure of a sample, whose value changes from sample to sample. Examples of statistics are X̄
and S 2 .
Bonus. State the Central Limit Theorem. You may use your own words.
The average (X̄) of a random sample of size n, coming from a population with mean µ and standard deviation σ, has an approximate normal distribution with mean µ and standard deviation
√
σ/ n, when n is sufficiently large.