# Exam1 review

1.
(Section 6.1) Normal distribution
with and
(a)

(b)
(c)
The number of days in which 75% of all accounts are above is the 25th percentile.

-5.3959 = X - 28
X = 22.6 days

2.
(Section 6.5) Sampling distribution of the mean
with , , and n = 20
(a)
(b)

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
(c)
np = 10(.9) = 9

4.
(Section 6.5) Sampling distribution of the mean
with , , n = 50
(a)

(b)
The central limit theorem

5.
(Section 6.1) Normal distribution
with ,
(a)
(b)

(c)
For IQR = Q3 - Q1, we need Q1, the 25th percentile, and Q3, the 75th percentile.

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

6.
(Section 6.5) Sampling distribution of the mean
with , , n = 25

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)
(b)

.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)

(b)

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)

(b)

12.
(Section 3.2) Central tendency/Variation
(a)
, 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)

13.
(Section 3.4) Empirical rule
The empirical rule states that 95% of the data falls within 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)
(b)
P(X = 0) = .00032
(c)
np = 5(.8) = 4

15.
(Section 6.1) Normal distribution
with ,
(a)

P(-1.25 < Z < 2.5) = .88814048
(b)
(c)
"Faster" means less time. 10% have quicker (smaller) times, and 90% have slower (larger) times. This is the 10th percentile.

16.
(Section 4.5) Binomial distribution
with n = 20, p = .04
(a)
P(X = 0) = .442002
(b)

17.
(Section 6.5) Sampling distribution of the mean
with , , n = 100
Note: If students buy tickets (or less) on the average, then the 250 tickets would be enough for for the 100 students.

18.
(Section 6.1) Normal distribution
with ,
(a)
If 75% of all workers earn more than you, then your wage is the 25th percentile.
per hour
(b)
(c)

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.