A certain group of test subjects had pulse rates with a mean of
77.6 beats per minute and a standard deviation of 10.2 beats pr minute.
Use the range rule of thumb to identify the limits separating values that are significantly low or significantly high.
Is a pulse rate of 68.0 beats per minute significantly low or significantly high?
 
 
Significantly low values are 57.2 beats per minute or lower. (Type an integer or a decimal. Do not round.)
 
Significantly high values are 98.0 beats per minute or higher. (Type an integer or a decimal. Do not round.)
 
Is a pulse rate of 68.0 beats per minute significantly low or significantly high?

Significantly high, because it is greater than two standard deviations above the mean.
Neither, because it is within two standard deviations of the mean.
Significantly low, because it is less than two standard deviations below the mean.
It is impossible to determine with the information given.
 
77.6 – 2 · 10.2 = 57.2
68.0 + 2 · 10.2 = 98.0
 

Class width is found by __________.     Choose the correct answer below.
 
subtracting a lower-class limit from the next consecutive lower-class limit.
adding the lower-class limit to the upper-class limit and dividing by two.
subtracting the smallest value in the data set from the largest value in the data set.
subtracting the lower limit of one class from the upper limit of that same class.
 

Below are the jersey numbers of 11 players randomly selected from a football team.
Find the range, variance, and standard deviation for the given sample data. What do the results tell us?
 
20       58       64       78      30       42       52      43       72       21      75
 
Range =     58                                        (Round to one decimal place as needed.)
Sample standard deviation =     21     (Round to one decimal place as needed.)
Sample variance =     400.8                 (Round to one decimal place as needed)
 
What do the results tell us?
Jersey numbers are nominal data that are just replacements for names, so the resulting statistics are meaningless.
The sample standard deviation is too large in comparison to the range.
Jersey numbers on a football team do not vary as much as expected.
Jersey numbers on a football team vary much more than expected.
 
Solution:
Range = highest minus lowest number in the data set.
78 – 20 = 58
 
Sample Standard Deviation: TI 84 Plus
  Steps

1st.     STAT
2nd.     EDIT
3rd.     ENTER DATA SET
4th.     STAT
5th.     CALC
6th.    Var Stats
7th.    Calculate

Sx = 20.99696948 (round to 21)
Sample variance = Standard Deviation² (Not the Sample Standard Deviation & do not round)
σx = 20.019824892² = 400.79339

The table below shows the frequency distribution of the rainfall on 52 consecutive
Tuesdays in a certain city use the frequency distribution to construct a histogram.
Do the data appear to have a distribution that is approximately normal?
Do the data appear to have a distribution that is approximately normal?
 
No, it is approximately uniform.
Yes, it is approximately normal.
No, it has no obvious maximum.
No, it is not symmetric.
 
Explanation:
Try and follow the frequency list I plugged in to the graph below!

Among fatal plane crashes that occurred during the past 60 years, 143 were due to pilot error.
72 were due to other human error, 511 were due to weather,
339 were due to mechanical problems, and 669 were due to sabotage.
 
Construct the relative frequency distribution.
What is the most serious threat to aviation safety, and can anything be done about it?
 
Complete the relative frequency distribution below Cause Pilot error.
 

 
What is the most serious threat to aviation safety, and can anything be done about it?
 
Weather is the most serious threat to aviation safety. Weather monitoring systems could be improved.
Mechanical problems are the most serious threat to aviation safety. New planes could be better engineered.
Pilot error is the most serious threat to aviation safety Pilots could be better trained.
Sabotage is the most serious threat to aviation safety. Airport security could be increased.
 
Explanation
First add the total number of fatal crash victims.
 
143 + 72 + 511 + 339 + 669 =      1734
 
Then divide the total number by each type and multiple by 100 to get the percentage.
 
Pilot Error: 143 ÷ 1739 = 0.0824682814 x 100 =                       8.2%
Human Error: 72 ÷ 1739 = 0.041522 x 100 =                             4.2%
Weather: 511 ÷ 1734 = 0.29469 · 100 =                                       29.5%
Mechanical Problems:  339 ÷ 1734 = 0.195501 · 100 =           19.6%
Sabotage: 669 ÷ 1734 = 0.3858131 · 100 =                                 38.6%

 
If your score on your next statistics test is converted to a z score, which of these z scores would you prefer:
-2.00, -1.00, 0, 1.00, 2.00? Why?
 
The z score of 2.00 is most preferable because it is 2.00 standard deviations above the mean and would correspond to the
highest of the five different possible test scores.
 
The z score of 0 is most preferable because it corresponds to a test score equal to the mean.
 
The z score of -1.00 is most preferable because it is 1.00 standard deviation below the mean and would
correspond to an above average test score.
 
The z score of -2.00 is most preferable because it is 2_00 standard deviations below the mean and would
correspond to the highest of the five different possible test scores.
 
The z score of 1.00 is most preferable because it is 1.00 standard deviation above the mean and would
correspond to an above average test score.

 
Which of the following is NOT a value in the 5-numers summary?
Choose the correct answer below.
 
Q₁
minimum
maximum
mean

 
The measure of center that is the value that occurs with the greatest frequency is the _____.
 
mode

 
A random sample of 10 subjects have weights with a standard deviation of 1098 kg.
What is the variance of their weights? Be sure to include the units with the result.
The variance of the sample data is 102.2081 kg².
(Round to four decimals as needed.)
 
Explanation
10.1098² = 102.2081 kg²

 
Below are the jersey numbers of 11 players randomly selected from a football team.
Find the range, variance, and standard deviation for the given sample data.
 
1          68       66      60      43       70      93       22       64       83       42
 
Range = 92 (Round to one decimal place as needed.)
Sample standard deviation = 26.8 (Round to one decimal place as needed.)
Sample variance = 716.3 (Round to one decimal place as needed.)
 
Explanation
 
Range = the highest minus the lowest number
93 – 1 = 92
Sample standard deviation
TI84 Plus Steps

 
STAT | EDIT | L1 (enter data set) | STAT  | CALC |  Vars Stats  | Calculate
 
(Please note: List must contain L1 & FreqList must be blank)
Remove L2 from the FreqList.
Just to be sure, go to: STAT | CALC | 1-Var Stats | FreqList (should not have L2 to the right of it).
 
Should Look Like This

If Freqlist has L2 next to it, then select the L2 and hit DEL to delete it.
(To put it back, hit 2^ND and the number 2).
 
TI84 Plus Steps

STAT | EDIT | L1 enter data set | STAT | CALC | Vars Stats | Calculate |
Sx = 26.7629323 (rounded to 26.8)
Sample Variance = Sample standard deviation² (do not round)
Sample Variance = Sx² = 26.7629323² = 716.2545453 (round to 716.3)

In a recent awards ceremony, the age of the winner for best actor was 37,
and the age of the winner for best actress was 48.
For all best actors, the mean age is 47.5 years, and the standard deviation is 7.3 years.
For all best actresses, the mean age is 35.2 years, and the standard deviation is 12.4 years.
(All ages are determined at the time of the awards ceremony.) Relative to their genders, had the
more extreme age when winning the award. the actor or the actress?
 
Since the z score for the actor is z = -1.44 and the z score for the actress is z = 1.03, the actress had the more age.
(Round to two decimal places.)
 
Explanation
 
Actor: 37 – 47.5 ÷ 7.3 = -1.4383556164 (round to -1.44)
Actress: 48 – 35.2 ÷ 12.4 = 1.032258065 (round to 1.3)
1.3 is the more extreme age.

 

 
Find the standard deviation, s, of sample data summarized in the frequency distribution table below by using the formula below,
where x represents the class midpoint, f represents the class frequency, and n represents the total number of sample values.
Use the calculator to find the standard deviation.
 
 
(Round to one decimal place as needed.)
 
Calculate the intervals like below…
20 + 26 ÷ 2 = 23
27 + 33 ÷ 2 = 30
34 + 40 ÷ 2 = 37
41 + 47 ÷ 2 = 44
48 + 54 ÷ 2 = 51
55 + 61 ÷ 2 = 58
62 + 68 ÷ 2 = 65
 
Sample standard deviation
TI84 Plus

 
Please note: List must contain L1 & L2
Just to be sure, go to: STAT | CALC | 1-Var Stats | FreqList
Should Look Like This, if not hit 2^ND and the number 2

 
Sample standard deviation
TI84 Plus
Steps
 
STAT
EDIT
L1 & L2
Enter Data Set
STAT
CALC
Vars Stats
Calculate
 
Data Set                    
Interval                Frequency (data given on the bottom of chart)
     L1                            L2
------------------------------------------------------------
     23                            2
     30                            2
     37                            2
     44                            2
     51                            13
     58                            35
     65                            29
 
Sx = 9.301170072 (rounded to 9.3)
 
You will need to use L2 in the FreqList.
Just to be sure, go to:
STAT
CALC
1-Var Stats and FreqList
hit 2^ND and the number 2.
It should appear like below!