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Exam1 review

Answers:

1.
(Section 6.1) Normal distribution
with $\mu = 28$ and $\sigma = 8$
(a)
$P(20 \leq X \leq 40) =
P(\frac{20 - 28}{8} \leq Z \leq \frac{40 - 28}{8}) =$
$P(-1 \leq Z \leq 1.5) = .774538$
(b)
$P(X \leq 30) = P(Z \leq \frac{30 - 28}{8}) =
P(Z \leq 0.25) = .59870627$
(c)
The number of days in which 75% of all accounts are above is the 25th percentile.
$-0.67448975 = \frac{X - 28}{8}$
-5.3959 = X - 28
X = 22.6 days

2.
(Section 6.5) Sampling distribution of the mean
with $\mu = 5$, $\sigma = 0.5$, and n = 20
(a)
$P(\overline{X} \geq 4.7) = P(Z \geq \frac{4.7 - 5}{0.5/\sqrt{20}}) =
P(Z \geq -2.68328) = .996355$
(b)
$P(\overline{X} \leq 4.8) = P(Z \leq \frac{4.8 - 5}{0.5/\sqrt{20}}) =
P(Z \leq -1.78885) = .03681908$

3.
(Section 4.5) Binomial distribution
with n = 10, p = .9
(a)
P(X = 9) = 0.38742
(b)
Not operating properly means we can just re-define the probability of success
to be p = .10. "No more than one" means 0 or 1, so
$P(X \leq 1) = .7361$
(c)
np = 10(.9) = 9

4.
(Section 6.5) Sampling distribution of the mean
with $\mu = 18$, $\sigma = 4$, n = 50
(a)
$P(16 \leq \overline{X} \leq 17) =
P(\frac{16 - 18}{4/\sqrt{50}} \leq Z \leq \frac{17 - 18}{4/\sqrt{50}}) =$
$P(-3.5356 \leq Z \leq -1.7678) = .03834637$
(b)
The central limit theorem

5.
(Section 6.1) Normal distribution
with $\mu = 80$, $\sigma = 10$
(a)
$P(X \leq 60) = P(Z \leq \frac{60 - 80}{10}) =
P(Z \leq -2) = .0227501$
(b)
$P(60 \leq X \leq 75) =
P(\frac{60 - 80}{10} \leq Z < \frac{75 - 80}{10}) =$
$P(-2 \leq Z \leq -0.5) = .28578747$
(c)
For IQR = Q3 - Q1, we need Q1, the 25th percentile, and Q3, the 75th percentile.
$-0.67448975 = \frac{X - 80}{10}$
$-6.7448975 = X - 80 \hspace{.3in} \Rightarrow \hspace{.3in}
X = 73.26 = Q_{1}$

$0.67448975 = \frac{X - 80}{10}$
$6.7448975 = X - 80 \hspace{.3in} \Rightarrow \hspace{.3in}
X = 86.75 = Q_{3}$

IQR = Q3 - Q1 = 86.75 - 73.26 = 13.49 minutes

6.
(Section 6.5) Sampling distribution of the mean
with $\mu = 8.5$, $\sigma = 2.5$, n = 25
$P(\overline{X} \geq 10) =
P(Z \geq \frac{10 - 8.5}{2.5/\sqrt{25}}) =
P(Z \geq 3) = .00134997$
Note that we must assume that our "symmetric" distribution is approximately normal,
since n is only 25.

7.
(Section 4.5) Binomial distribution
with n = 8, p = .1
(a)
$P(X \leq 1) = .8131$
(b)
$P(2 \leq X \leq 3) = P(X = 2) + P(X = 3) =$
.1488 + .03307 = .18187
(c)
P(X = 8) = .00000001
(d)
np = 8(.1) = 0.8

8.
(Section 6.6) Sampling distribution of the proportion
with n = 125, p = .10
Note: This problem could also be solved with the binomial distribution.
(a)
$p_{s} = \frac{15}{125} = .12$
$P(p_{s} \geq .12) =
P(Z \geq \frac{.12 - .10}{\sqrt{\frac{.10(.90)}{125}}}) =$
$P(Z \geq 0.745356) = .22802820$
(b)
$p_{s} = \frac{10}{125} = .08$
$P(p_{s} \leq .08) =
P(Z \leq \frac{.08 - .10}{\sqrt{\frac{.10(.90)}{125}}}) =$
$P(Z \leq -0.745356) = .22802820$

9.
(Section 1.7) Types of data
(a)
Numerical, discrete
(b)
Categorical
(c)
Categorical
(d)
Numerical, continuous
(e)
Numerical, continuous
(f)
Categorical ("Yes" or "No")

10.
(Section 2.1) Stem-and-leaf plots
*Also see section 3.2 (shape, mode)
(a)
The shape is approximately symmetric.
(b)
mode = 21 (it appears 3 times)

11.
(Section 6.6) Sampling distribution of the proportion
with n = 50, p = .70
Note: This problem could be solved with the binomial distribution as well.
(a)
$P(p_{s} \geq .80) =
P(Z \geq \frac{.80 - .70}{\sqrt{\frac{.7(.3)}{50}}}) =$
$P(Z \geq 1.54303) = .0614113$
(b)
$P(p_{s} \geq .50) =
P(Z \geq \frac{.50 - .70}{\sqrt{\frac{.7(.3)}{50}}}) =$
$P(Z \geq -3.08607) = .99898582$
12.
(Section 3.2) Central tendency/Variation
(a)
$\overline{X} = 4$, Q2 = 6
(b)
s = 6.15088, s2 = (6.15088)2 = 37.83332
(c)
range = max - min = 12 - (-8) = 20, IQR = Q3 - Q1 = 8.5 - (-1) = 9.5
(d)
$midhinge = \frac{Q_{1} + Q_{3}}{2} = \frac{-1 + 8.5}{2} = 3.75,$
$midrange = \frac{min + max}{2} = \frac{-8 + 12}{2} = 2$

13.
(Section 3.4) Empirical rule
The empirical rule states that 95% of the data falls within $\pm$ 2 standard deviations of the mean. Therefore, the distance from 15 to 35 (which is 35 - 15 = 20) should be 4 standard deviations. So we have 4s = 20; s = 5.

14.
(Section 4.5) Binomial distribution
with n = 5, p =.80
(a)
$P(X \geq 4) = P(X = 4) + P(X = 5) = .4096 + .32768 = .73728$
(b)
P(X = 0) = .00032
(c)
np = 5(.8) = 4

15.
(Section 6.1) Normal distribution
with $\mu = 20$, $\sigma = 4$
(a)
$P(15 \leq X \leq 30) =
P(\frac{15 - 20}{4} \leq Z \leq \frac{30 - 20}{4}) =$
P(-1.25 < Z < 2.5) = .88814048
(b)
$P(X \geq 30) =
P(Z \geq \frac{30 - 20}{4}) =
P(Z \geq 2.5) = .00620968$
(c)
"Faster" means less time. 10% have quicker (smaller) times, and 90% have slower (larger) times. This is the 10th percentile.
$-1.28155 = \frac{X - 20}{4}
\hspace{.3in} \Rightarrow \hspace{.3in}
X = 14.8738$

16.
(Section 4.5) Binomial distribution
with n = 20, p = .04
(a)
P(X = 0) = .442002
(b)
$P(X \geq 1) = 1 - P(X = 0) = 1 - .442002 = .557998$

17.
(Section 6.5) Sampling distribution of the mean
with $\mu = 2.4$, $\sigma = 2$, n = 100
Note: If students buy $\overline{X} = \frac{250}{100} = 2.5$ tickets (or less) on the average, then the 250 tickets would be enough for for the 100 students.
$P(\overline{X} \leq 2.5) =
P(Z \leq \frac{2.5 - 2.4}{2/\sqrt{100}}) =
P(Z \leq 0.5) = .69146247$

18.
(Section 6.1) Normal distribution
with $\mu = 11.90$, $\sigma = 0.40$
(a)
If 75% of all workers earn more than you, then your wage is the 25th percentile.
$-0.67449 = \frac{X - 11.90}{0.40}
\hspace{.3in} \Rightarrow \hspace{.3in}
X = \$11.63$ per hour
(b)
$P(X \leq 11) = P(Z \leq \frac{11 - 11.90}{0.40}) =
P(Z \leq - 2.25) = .01222443$
(c)
$P(12 \leq X \leq 13) =
P(\frac{12 - 11.90}{0.40} \leq Z \leq \frac{13 - 11.90}{0.40}) =$
$P(0.25 \leq Z \leq 2.75) = .39831391$

19.
(Section 6.2) Normal probability plots
(a)
Based on the shape of the boxplot and histogram, the data looks non-normal. Both plots are highly right-skewed.
(b)
The normal probability plot is far from a straight line. This data does NOT look normal.



 
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2001-01-26