Stat 216 Exam One Review Questions

 

  1. General Hospital's patient account division has compiled data on the age of accounts receivables. The data collected indicate that the age of the accounts follows a normal distribution with mean 28 days and standard deviation 8 days.
    1. What proportion of the accounts are between 20 and 40 days old?
    2. What proportion of the accounts are less than 30 days old?
    3. What is the number of days in which 75% of all accounts are above?
    [Go to answer]

     

  2. Reclaimed phospate land in Polk County, Florida, has been found to emit a higher mean radiation level than other nonmining land in the county. Suppose that the radiation level for the reclaimed land has a distribution with mean 5.0 working levels (WL) and a standard deviation of 0.5 WL. Suppose further that 20 houses built on reclaimed land are randomly selected and the radiation level is measured in each.
    1. What is the probability that the sample mean for the 20 houses exceeds 4.7 WL?
    2. What is the probability that the sample mean is less than 4.8 WL?
    [Go to answer]

     

  3. Based on past experience, the main printer in a university computer center is operating properly 90% of the time. Suppose inspections are made at 10 randomly selected times.
    1. What is the probability that the main printer is operating properly for exactly 9 of the inspections?
    2. What is the probability that the main printer is not operating properly no more than 1 inspection?
    3. What is the expected number of inspections in which the main printer is operating properly?
    [Go to answer]

     

  4. The length of time of long-distance telephone calls has mean of 18 minutes and standard deviation of 4 minutes. Suppose a sample of 50 telephone calls is used to reflect on the population of all long-distance calls.
    1. What is the chance that the average of the 50 calls is between 16 and 17 minutes?
    2. What theorem do we need in order to solve (a.)?
    [Go to answer]

     

  5. The time required to complete a final examination in a particular college course is normally distributed, with mean of 80 minutes and a standard deviation of 10 minutes. Answer the following questions.
    1. What is the probability of completing the exam in one hour or less?
    2. What is the probability a student will complete the exam in a time between 60 and 75 minutes?
    3. What is the interquartile range for completion times?
    [Go to answer]

     

  6. Assume that the dividends of electric utility stocks as of a given date have a symmetric distribution with mean of 8.5 percent and standard deviation of 2.5 percent. Find the probability that the average dividend of 25 such stocks will exceed 10 percent.
    [Go to answer]

     

  7. The probability that a patient fails to recover from a particular operation is 0.1. Suppose that eight patients having this operation are selected at random. Answer the following questions.
    1. What is the probability that at most one patient will not recover?
    2. What is the probability that at least 2 but no more than 3 patients will not recover?
    3. What is the probability that all patients will not recover?
    4. What is the expected number of patients that will not recover?
    [Go to answer]

     

  8. An insurance company states that 10% of all fire insurance claims are fraudulent. Suppose the company is correct, and that it receives 125 claims.
    1. What's the probability that at least 15 claims are fraudulent?
    2. What's the probability that less than 10 claims are fraudulent?
    [Go to answer]

     

  9. What type of data would be collected by the following survey questions?
    1. "How many pairs of shoes do you own?"
    2. "What color are your eyes?"
    3. "Which brand of soft drink do you prefer?"
    4. "What is the circumference of that tree?"
    5. "What is your GPA?"
    6. "Are you planning to vote this year?"
    [Go to answer]

     

  10. A random sample of 33 price-earnings ratios for a set of stocks whose prices are quoted by NASDAQ is displayed in the following stem-and-leaf plot.
         Stem-and-Leaf Display			
    			
         Stem unit:	10		
    			
         0   4		
         1   0 5 6 8 8 9		
         2   0 1 1 1 2 3 4 5 5 6 7 8 8 9 9		
         3   0 1 2 3 4 4 5 7		
         4   1 7		
         5  		
         6   2		
    
    1. What is the shape of this data?
    2. What is the mode of this dataset?
    [Go to answer]

     

  11. Seventy percent of small businesses experience cash flow problems during their first year of operation. A consultant takes a random sample of 50 small businesses that have been in business for one year.
    1. What is the probability that more than 80% of the sample have experienced cash flow problems?
    2. What is the probability that more than half of the sample has had cash flow problems?
    [Go to answer]

     

  12. Consider the following sample of data. Obtain the "1-Var Stats" from the TI-83 in order to answer the following questions.
      7, -5, -8, 7, 9, 12, 0, 3, 11, 8, 6, -2, 4
      
    1. What are the mean and median for this dataset?
    2. What are the standard deviation and variance?
    3. What are the range and interquartile range?
    4. What are the midhinge and midrange?
    [Go to answer]

     

  13. Consider the following boxplot. Suppose 95% of the data falls between 15 and 35. Based on the empirical rule, what is the standard deviation of this sample of data? Boxplot
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  14. A company estimates that there is an 80% chance of an order arriving on time from a supplier. Suppose 5 orders are placed this week.
    1. What is the probability that at least 4 orders arrive on time?
    2. What is the probability that none of the orders arrive on time?
    3. How many orders would you expect to arrive on time?
    [Go to answer]

     

  15. The reaction time to a certain psychological experiment is considered to be normally distributed with a mean of 20 seconds and a standard deviation of 4 seconds.
    1. What proportion of subjects take between 15 and 30 seconds to react?
    2. What proportion of subjects take longer than 30 seconds to react?
    3. What is the reaction time such that only 10% of subjects are faster?
    [Go to answer]

     

  16. A natural gas exploration company averages four strikes (that is, natural gas is found) per 100 holes drilled.
    1. If 20 holes are to be drilled, what is the probability that no strikes will be made?
    2. What is the probability that at least one strike will be made?
    [Go to answer]

     

  17. Suppose you are in charge of student ticket sales for a college football team. From past experience, you know that the number of tickets purchased by a student standing in line at the ticket window has a distribution with mean 2.4 and standard deviation 2.0. For today's game, there are 100 students standing in line to purchase tickets. If only 250 tickets remain, what is the probability that all 100 students will be able to purchase the tickets they desire? (Hint: 250 tickets for 100 students would be how many on average for each student?)
    [Go to answer]

     

  18. Wages for workers in a particular industry average $11.90 per hour with a standard deviation of 40 cents. The wages are considered to be normally distributed.
    1. Suppose you are employed in this industry. What would your wage have to be if 75% of all workers earn more than you?
    2. What proportion of workers receive wages less than $11 per hour?
    3. What proportion of workers make between $12 and $13 per hour?
    [Go to answer]

     

  19. Researchers have developed sophisticated intrusion-detection algorithms to protect the security of computer-based systems. These algorithms use principles of statistics to identify unusual or unexpected data, i.e., "intruders". One popular intrusion-detection system assumes the data being monitored are normally distributed. Consider the following data on input-output (I/O) units utilized by a sample of 44 users of a system.
     15   5   2  17   4   3   1   1   0   0   0   0
      0   0   0  20   9   0   0   0   1   6   1   3
      1   0   6   0   0   0   0   1  14   0   7   0
      2   9   4   0   0   0   9  10
    
    Use your TI-83 to decide whether or not the data appear to be approximately normal:
    1. Obtain a boxplot and histogram of this data. Based on the shape, does it look like this sample comes from a normally distributed population?
    2. Obtain a normal probability plot. What is your conclusion?
    [Go to answer]

Answers:


1.
(Section 6.1) Normal distribution
with $\mu = 28$ and $\sigma = 8$
(a)
$P(20 \leq X \leq 40) =
P(\frac{20 - 28}{8} \leq Z \leq \frac{40 - 28}{8}) =$
$P(-1 \leq Z \leq 1.5) = .774538$

normal curve       TI-83 screen

(b)
$P(X \leq 30) = P(Z \leq \frac{30 - 28}{8}) =
P(Z \leq 0.25) = .59870627$

normal curve       TI-83 screen

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

normal curve

normal curve

$-0.67448975 = \frac{X - 28}{8}$
-5.3959 = X - 28
X = 22.6 days

TI-83 screen TI-83 screen

2.
(Section 6.5) Sampling distribution of the mean
with $\mu = 5$, $\sigma = 0.5$, and n = 20
(a)
$P(\overline{X} \geq 4.7) = P(Z \geq \frac{4.7 - 5}{0.5/\sqrt{20}}) =
P(Z \geq -2.68328) = .9963548$

normal curve       TI-83 screen

(b)
$P(\overline{X} \leq 4.8) = P(Z \leq \frac{4.8 - 5}{0.5/\sqrt{20}}) =
P(Z \leq -1.78885) = .03681908$

normal curve       TI-83 screen

3.
(Section 4.5) Binomial distribution
with n = 10, p = .9
(a)
P(X = 9) = 0.38742

TI-83 screen TI-83 screen

TI-83 screen TI-83 screen

(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)
$P(16 \leq \overline{X} \leq 17) =
P(\frac{16 - 18}{4/\sqrt{50}} \leq Z \leq \frac{17 - 18}{4/\sqrt{50}}) =$
$P(-3.5356 \leq Z \leq -1.7678) = .03834637$

normal curve       TI-83 screen

(b)
The central limit theorem

5.
(Section 6.1) Normal distribution
with $\mu = 80$, $\sigma = 10$
(a)
$P(X \leq 60) = P(Z \leq \frac{60 - 80}{10}) =
P(Z \leq -2) = .0227501$

normal curve       TI-83 screen

(b)
$P(60 \leq X \leq 75) =
P(\frac{60 - 80}{10} \leq Z < \frac{75 - 80}{10}) =$
$P(-2 \leq Z \leq -0.5) = .28578747$

normal curve       TI-83 screen

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

normal curve

normal curve

$-0.67448975 = \frac{X - 80}{10}$
$-6.7448975 = X - 80 \hspace{.3in} \Rightarrow \hspace{.3in}
X = 73.26 = Q_{1}$

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

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

TI-83 screen

6.
(Section 6.5) Sampling distribution of the mean
with $\mu = 8.5$, $\sigma = 2.5$, n = 25
$P(\overline{X} \geq 10) =
P(Z \geq \frac{10 - 8.5}{2.5/\sqrt{25}}) =
P(Z \geq 3) = .00134997$

normal curve       TI-83 screen

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

TI-83 screen TI-83 screen

TI-83 screen

(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)
$p_{s} = \frac{15}{125} = .12$
$P(p_{s} \geq .12) =
P(Z \geq \frac{.12 - .10}{\sqrt{\frac{.10(.90)}{125}}}) =$
$P(Z \geq 0.745356) = .22802820$

normal curve       TI-83 screen

(b)
$p_{s} = \frac{10}{125} = .08$
$P(p_{s} \leq .08) =
P(Z \leq \frac{.08 - .10}{\sqrt{\frac{.10(.90)}{125}}}) =$
$P(Z \leq -0.745356) = .22802820$

normal curve       TI-83 screen

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)
$P(p_{s} \geq .80) =
P(Z \geq \frac{.80 - .70}{\sqrt{\frac{.7(.3)}{50}}}) =$
$P(Z \geq 1.54303) = .0614113$

normal curve       TI-83 screen

(b)
$P(p_{s} \geq .50) =
P(Z \geq \frac{.50 - .70}{\sqrt{\frac{.7(.3)}{50}}}) =$
$P(Z \geq -3.08607) = .99898582$

normal curve       TI-83 screen

12.
(Section 3.2) Central tendency/Variation

TI-83 screen TI-83 screen

TI-83 screen

(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

TI-83 screen TI-83 screen

(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

normal curve       TI-83 screen

(b)
$P(X \geq 30) =
P(Z \geq \frac{30 - 20}{4}) =
P(Z \geq 2.5) = .00620968$

normal curve       TI-83 screen

(c)
"Faster" means less time. 10% have quicker (smaller) times, and 90% have slower (larger) times. This is the 10th percentile.

normal curve

normal curve

$-1.28155 = \frac{X - 20}{4}
\hspace{.3in} \Rightarrow \hspace{.3in}
X = 14.8738$

TI-83 screen

16.
(Section 4.5) Binomial distribution
with n = 20, p = .04

TI-83 screen TI-83 screen

(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.
$P(\overline{X} \leq 2.5) =
P(Z \leq \frac{2.5 - 2.4}{2/\sqrt{100}}) =
P(Z \leq 0.5) = .69146247$

normal curve       TI-83 screen

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.

normal curve

normal curve

$-0.67449 = \frac{X - 11.90}{0.40}
\hspace{.3in} \Rightarrow \hspace{.3in}
X = \$11.63$ per hour

TI-83 screen

(b)
$P(X \leq 11) = P(Z \leq \frac{11 - 11.90}{0.40}) =
P(Z \leq - 2.25) = .01222443$

normal curve       TI-83 screen

(c)
$P(12 \leq X \leq 13) =
P(\frac{12 - 11.90}{0.40} \leq Z \leq \frac{13 - 11.90}{0.40}) =$
$P(0.25 \leq Z \leq 2.75) = .39831391$

normal curve       TI-83 screen

19.
(Section 6.2) Normal probability plots
(a)

TI-83 screen TI-83 screen

Based on the shape of the boxplot and histogram, the data looks non-normal. Both plots are highly right-skewed.
(b)

TI-83 screen

The normal probability plot is far from a straight line. This data does NOT look normal.