Question 1

Arithmetic mean = (28 + 32 + 28 + 30 + 34)/5 = 30.4

Median = 30

Mode = 28

Based on the above results, arithmetic mean would be the best measure of center because it lies approximately at the center of the five runs.

Question 2

Arithmetic mean = (31 + 29 + 31 + 29 + 31)/5 = 30.2

Median = 31

Mode = 31

Based on the above results, arithmetic mean would be the best measure of center because it lies approximately at the center of the five runs.

Question 3

Arithmetic mean = (29 + 32 + 28 + 32 + 30)/5 = 30.2

Median = 30

Mode = 32

Based on the above results, median would be the best measure of center because it lies at the center of the five runs.

Question 4

The smallest number is 101 and the largest number is 157. Therefore, the range is 157 – 101 = 56. To get the class width, divide 56 by 5. That is, . Rounding off to the nearest whole number, each class will have a width of 11. The classes will therefore be 101-112, 113-124, 125-136, 137-148 and 149-160

Class | Frequency | |

101-112 | 3 | 0.12 |

113-124 | 11 | 0.44 |

125-136 | 7 | 0.28 |

137-148 | 2 | 0.08 |

149-160 | 2 | 0.08 |

Question 5

Question 6

Question 7

The above distribution is skewed to the left meaning that it is not symmetric.

Question 8

Class | Frequency | Mid point (x) | f.x |

101-112 | 3 | 106.5 | 319.5 |

113-124 | 11 | 118.5 | 1303.5 |

125-136 | 7 | 130.5 | 913.5 |

137-148 | 2 | 142.5 | 285 |

149-160 | 2 | 154.5 | 309 |

25 | 3130.5 |

Sample mean =

Question 9

Range = the highest pay – the lowest pay = 1019 – 444 = 575

Question 10

Mean =

Variance =

Question 11

Standard deviation =

**Box-and-whisker plots**

Question 1

Min: 2

Q1: 6

Q2: 11

Q3: 13

Max: 18

Question 2

Ordering the data we have 4, 5, 7, 10, 10, 10, 11, 13, 13, 14, 17, 18, 19.

Min: 4

Q1= (7 + 10)/2 = 17/2 = 8.5

Q2 = (n + 1)/2 =14/2 = 7^{th} value = 11

Q3 = (14 + 17)/2 = 31/2 = 15.5

Max: 19

Outlier: none

Outlier: anything below Q1 – 1.5 IQR or anything above Q3 + 1.5 IQR

IQR = Q3 – Q1 = 15.5 – 8.5 = 7

Therefore, Q1 -1.5 IQR = 8.5 – (1.5*7) = 8.5 – 10.5 = -2

And Q3 + 1.5 IQR = 15.5 + 10.5 = 26

Given that no data lies below -2 and above 26 then there are no outliers.

Question 3

No, it is not common for a teenager to spend more than 1 hour getting ready for school. In addition, it is not common for a teenager to spend between 1 and 2 hours getting ready for a school dance. The reason is that in both cases majority of the students spend 1 hour or less getting ready for either school or school dance.

Question 4

Only 25 percent of the students spend at least 15 minutes getting ready for a school dance.

Question 5

True because Q3 – Q2 is approximately 25 percent

Question 6

It might not be possible to determine the number of girls that shop for clothes in September because the box plot does not provide such data. It only provides the amount of money they spend on clothes.

Question 7

The percentage of the girls that spent less than $85 is 25 because the lower quartile lies at $85.

Question 8

I would expect the mean number of dollar spent on clothes to be higher than the median. This is in relation to the fact that the box plot is skewed to the right. Therefore, it would be reasonable to expect the mean to be higher than the median.

**Stem and leaf diagrams**

Question 1

Ordering the data we have 8, 9, 11, 13, 17, 18, 18, 21, 22, 25, 27, 28, 29, 31, 32.

Stem | Leaf |

0 | 8, 9 |

1 | 1, 3, 7, 8, 8 |

2 | 1, 2, 5, 7, 8, 9 |

3 | 1, 2 |

Question 2

Range = 40 – 8 = 32

Question 3

Students with GPA of 2 and above are 12

Question 4

The mode is 14 and 25

Question 5

Median = (25 + 25)/2 = 25

Question 6

Based on the results provided, there are 18 students in Mr. Smith’s math class.

**Scatter plots & line graphs**

Question 1

The scatter plot shows a positive relationship between the number of hours students spend studying and their grades. This means that as the number of hours of studying increase, the grade improves. It also indicates that as the number of hours of studying decrease, the grade drops.

Question 2

a). Graph 3 shows no relationship between temperature and test score because the plots do not form a consistent pattern.

b). Graph 2 shows a positive correlation between temperature and score because as the temperature increases, the score increases as well. In addition, as the temperature decreases, the score decreases as well.

**Basic probability**

Question 1

This is an experimental probability because the probability provided is based on company’s records. If the records were not used, it would have been a theoretical one. However, given that company’s records were used, then it is an example of experimental probability.

Question 2

This is an example of a theoretical probability because the probability provided is based on expectations.

Question 3

The probability of selecting an even number between 1 and 15 is as it follows;

Even numbers between 1 and 15 are 2, 4, 6, 8, 10, 12, and 14. That is, they are 7. Therefore, the probability of selecting an even number between 1 and 15 is

Question 4

Numbers divisible by 4 between 1 and 50 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, and 48. This means that there are 12 numbers divisible by 4 between 1 and 50. Therefore, the probability of selecting one of these numbers is

Question 5

A standard deck has 52 cards. The 52 cards are divided into four groups of spades, hearts, diamonds and clubs. Each of these four groups has an ace. Therefore, there are 4 aces. Based on this fact, the probability of drawing an ace from a standard deck of playing cards is

Question 6

The bag contains 14 marbles (6 red, 3 green and 5 yellow). The probability of not selecting a green marble would be 1 – the probability of selecting a green marble. That is,

Question 7

A standard deck has 52 cards. The 52 cards are divided into four groups of spades, hearts, diamonds and clubs. Each of these four groups has a king. Therefore, there are 4 kings in a standard deck of cards. Based on this fact, the probability of not selecting a king from the standard deck would be

**Multiplication rule**

Question 1

For this event the probability of selecting an ace would be 4 out of 52 and the probability of selecting a jack would also be 4 out of 52. This being the case, the events would be independent because the first event would not affect the outcome of the second event in any way.

Question 2

For this event, the probability of selecting the first ball would be 1 out of 100, but the probability of selecting the second ball would be 1 out of 99. This means that the first event would affect the outcome of the second event. As a result, the two events would be dependent.

Question 3

In this case, the probability of getting a head throughout would be 1 out of 2. However, because the exercise would be repeated eight times, then the probability of getting a head every time would be (½)^{8} = 1/2^{8} = 1/256

Question 4

A standard deck of cards has 52 cards; 13 diamonds; 4 queens and 13 spades. Therefore, the probability of selecting a diamond with replacement would be 13/52. The probability of selecting a queen with replacement would be 4/52 and the probability of selecting a spade with replacement would be 13/52. In total, three cards are selected with replacement. Therefore, their probability would be

Question 5

A standard deck of cards has 52 cards, 1 queen of hearts and 13 hearts including the queen of hearts. Therefore, the probability of selecting the queen of hearts would be 1/52. The probability of selecting any heart after the queen of hearts has been selected would be 12/51 because one heart would be selected already and the number of cards would have reduced by one. In total, the probability of selecting the two cards would be

Question 6

There are 11 parts in the box

4 are defective and 7 are not defective

P (both parts are defective) =

Question 7

P (both parts not defective) =

Question 8

P (at least one defective) = [P (defective) and P (not defective)] + [P (not defective) and P (defective)]

=() + () =

**The addition rule**

Question 1

The events of selecting a male student and selecting a senior are not mutually exclusive because there are 325 males in the seniors. Consequently, in the process of selecting a male student you can select a senior student. In the process of selecting a senior student, you can also select a male student. Therefore, the two events are not mutually exclusive.

Question 2

P (male or senior) = P (male) + P (senior) – P (senior male only)

P (male) =

P (senior) =

P (senior male only) =

P (male or a senior) =

Question 3

Selecting a heart or 6

Mutually exclusive? | No |

Probability | P (selecting a 6) + P (selecting a heart) – P (6 and heart) = |

Question 4

Selecting a red suit or a jack

Mutually exclusive? | No |

Probability | P (a red suit) + P (jack) – P (a red suit and jack) = |

Question 5

Rolling a die: rolling a 6 or a number greater than 4

Mutually exclusive? | No |

Probability | P ( 5) + P (6) = |

Question 6

Rolling a 5 or an even number

Mutually exclusive? | Yes |

Probability | P (5) + P (2) + P (4) + P (6) = |

**Combinations & permutations**

Question 1

_{8}P_{3} =

Question 2

_{8}C_{3} =

Question 3

Given that the first digit cannot be zero, then the numbers can be arranged in _{9}P_{4} different ways. However, because the last digit should be even, we should subtract the arrangements that would end with odd numbers from _{9}P_{4}. In this case, the arrangements that would end with odd numbers would be 5!

9P4 =

Subtracting 5! From 3,024 we get (3024 – 120) = 2,904 codes

Question 4

Given that the horses cannot tie, then the 8 horses can finish the race in 8*7*6*5*4*3*2*1 ways. This equals 40,320 different ways.

Question 5

Smith has a total of 4 + 8 + 6 = 18 trees. Given that order would be insignificant in this case, then Smith can plant these trees in 18! different ways.

Question 6

The jury can be selected _{40}C_{12} different ways because order is important.

40C12 =

**Probability exam**

Question 1

In total, there were 2,466,000 degrees. Out of these degrees, 467,000 were master’s degree. Therefore, the probability of selecting a person with master’s degree is

Question 2

The females that earned master’s degree are 270,000. Therefore, the probability of selecting a female with master’s degree would be

Question 3

The males with associate’s degrees were 231,000. Therefore, the probability of selecting a male with associate’s degree would be

Question 4

The number of the people that did not earn doctorate is 632,000 + 1,322,000 + 467,000 = 2,421,000. Therefore, the probability of selecting a person that did not earn a doctorate degree is

Question 5

In this case, we have 26 letters and 10 digits. Then we have 26 + 10 = 36 letters and digits to arrange on the license plate. Given that only 4 characters should appear on the license plate, then we can have 36P4 possible license plates.

Question 6

The 15 swimmers can finish first, second and third in 15P3 different ways.

Question 7

In this case, order is an important aspect. Therefore, the editor can select the stories in 17C4 different ways. That is

Question 8

We have 11 characters in *mathematics*. However, there are 2 A’s, 2 M’s, 2 T’s. Therefore, if we are to arrange these characters in different ways, the many times that A’s would appear together, the arrangement would not be distinguishable. The same case would apply to M’s and T’s. As a result, if we are to arrange these characters, we would only have 8! distinguishable outcomes.

Question 9

These events would be independent because the sixth toss does not depend on the other five toss.

Question 10

The two events are not mutually exclusive because a king card is also a face card. As a result, the outcome of event A will automatically affect the outcome of event B.

Probability distributions

Question 1

X | 1 | 2 | 3 | 4 | 5 | 6 |

P(x) | 0.02 | 0.34 | 0.18 | 0.26 | 0.05 | 0.15 |

Probability distribution | Yes |

Why or why not? | Because the sum of P(x) is equal to one |

Question 2

X | 1 | 2 | 3 | 4 |

P(x) | 0.31 | 0.08 | 0.48 | 0.16 |

Probability distribution? | No |

Why or why not? | Because the sum of P(x) is equal to 1.03 and not equal to one |

Question 3

X | 1 | 2 | 3 | 4 | 5 |

P(x) | ¾ | 5/4 | ½ | ¼ | -7/4 |

Probability distribution? | No |

Why or why not? | Because P(X=5) is negative. Therefore, even if the sum of P(x) is equal to 4/4 which is one, the condition that p(x) for any real number should be non-negative is not met. |

Question 4

Pets | 0 | 1 | 2 | 3 | 4 | 5 |

Households | 1491/2175 = 0.6855 | 425/2175 = 0.1954 | 168/2175 = 0.0772 | 48/2175 = 0.0221 | 29/2175 = 0.0133 | 14/2175 = 0.0065 |

Question 5

Computers | 0 | 1 | 2 | 3 |

Households | 300/695 = 0.4317 | 280/695 = 0.4029 | 95/695 = 0.1367 | 20/695 = 0.0287 |

**More probability distributions**

Question 1

Mean =

=0.1954 + 0.1544 + 0.0663 + 0.0532 + 0.0325 = 0.5018

Expected value = mean = 0.5018

Variance =

= (0-0.5018)^{2}*0.6855 + (1-0.5018)^{2} *0.1954 + (2-0.5018)^{2}*0.0772 + (3-0.5018)^{2}*0.0221 + (4-0.5018)^{2}*0.0133 + (1-0.5018)^{2}*0.0065

= 0.172611 + 0.048499 + 0.173283 + 0.137926 + 0.162757 + 0.13152

= 0.826597

Standard deviation =

Question 2

Mean =

=0.4029 + 0.2734 + 0.0861

=0.7624

Expected value = mean = 0.7624

Variance =

=0.250927 + 0.022745 + 0.209377 + 0.143697

= 0.626746

Standard deviation =

**Binomial distributions**

Question 1

This is a binomial experiment because a person that enters the clothing store can either buy or not buy something from the store. The following are the values for the experiment.

n = 16;

p = 0.28;

q = 0.72;

Question 2

This is not a binomial experiment because there are three possible outcomes. The three possible outcomes are approve, disapprove and no opinion. This being the case, the experiment cannot be binomial.

Question 3

n = 5, p = ¼, q = ¾

a). P (X=3) =

=

b). P (X=5) =

=

=

Question 4

P = 0.54, q = 0.46, n = 10

**More binomial distributions**

Question 1

n = 100; p = 0.6; q = 0.4

Mean = np = 100*0.6 = 60

Variance = npq = 100*0.6*0.4 = 24

Standard deviation =

Question 2

n = 46; p = 0.18; q = 0.82

Mean = np = 46*0.18 = 8.28

Variance = npq = 46*0.18*0.82 = 6.7896

Standard deviation =

Question 3

P = 0.63, q = 0.37, n = 5

Mean = np = 5*0.63 = 3.15

Variance = npq = 5*0.63*0.37 = 1.1655

Standard deviation =

The results show that about three teenagers have jobs. The standard deviation of these teenagers is 1.07958 and their variance is 1.1655.

Question 4

n = 20; p = 0.38; q = 0.62

Mean = np = 20*0.38 = 7.6

Variance = npq = 20*0.38*0.62 = 4.712

Standard deviation =

The results show that about eight people of the interviewees had type O^{+} blood. The standard deviation of these interviewees was 2.1707 and their variance was 4.712.

**The normal distribution**

Question 1

For this case, it is almost possible that a volume of 11.4 oz will be found in the population of the soda. This is in relation to the fact that upon standardizing the sample, we get a z-score of -3. This indicates 99.7 percent likelihood.

Question 2

P (150≤X≤200) = P

= P (1≤Z≤3)

= 0.1587 – 0.0013

= 0.1574

This means that the likelihood that the values will be between 150 and 200 is 0.1574 or 15.74 percent. This is a small percentage indicating that the values are unlikely to lie in the range.

Question 3

P (48<X<61) = P

= P (-1<Z<1)

= 0.3413 + 0.3413

= 0.6826

This shows that the likelihood of a male dog of the same breed lying in the range between 48 lb and 61 lb is 68.26 percent. This indicates a high likelihood.

Question 4

P (X=0.004) = P

= P (Z=3)

= 0.5 + (0.5 – 0.0013)

= 0.9987

This shows that a value of 0.004 is very common in the population.

**Finding probabilities**

Question 1

Mean = 124, standard deviation = 74

P (X>198) = P

= P (Z>1)

= 0.5 – 0.34

= 0.16

Question 2

P (23<X<337) = P

= P (0<Z<2)

= 0.5 – 0.475

= 0.025

Question 3

P (X>310) = P

= P (Z>3)

= 0.5 – 0.4985

= 0.0015

Question 4

P (-286<X<470) = P

= P (-2<Z<2)

= 95%

**Finding values**

Question 1

A z-score of -3.4 means that 99.97 percent of the distribution lies above -3.4. It also means that only 0.03 percent of the distribution lies below -3.4.

Question 2

X – 70 = -1.82*62

X = 70 – 112.84

X = -42.84

Question 3

µ = 60, δ = 98

Z-score =

Question 4

P (Z<2.02) = 0.5 + (0.5 – 0.0217)

= 0.5 + 0.4783

= 0.9783

Question 5

P (X>-1.97) = 0.5 + (0.5 – 0.0244)

= 0.5 + 0.4756

= 0.9756

**The central limit theorem**

Question 1

P (Z=0.0017) = 1 – 0.0017

= 0.9983

Question 2

P (X<33.5) = P

= P (Z<-0.3)

= 0.5 – 0.3821

= 0.1179

Question 3

P (X<30.92) = P

= P (Z<0.092)

= 0.5 + 0.4641

= 0.9641