Back to the class
Section 2.5 #12: Consider the data:
36  41  39  47  15  48  34  28  25  28  19  18  50  27  53

(a) find the quartiles,
(b) find the interquartile range,
(c) identify any outliers.
Solution: For (a), first recall that the second quartile, $Q_2$, is given by the median of the data. The first quartile $Q_1$ is the median of all the data to the left of $Q_2$ and the third quartile $Q_3$ is the median of all the data to the right of $Q_2$. We find $Q_2$ first. To do so, first order the data:
15 18 19 25 27 28 28 34 36 39 41 47 48 50 53
From this we see that the second quartile (i.e. the median) is $Q_2=34$. To find $Q_1$ we find the median of the data left of $Q_2$, i.e. the median of the following data:
15 18 19 25 27 28 28
which is $Q_1=25$. To find $Q_3$ we find the median of the data to the right of $Q_2$, i.e. the median of the following data:
36 39 41 47 48 50 53
which is $Q_3=47$.

For (b), the interquartile range ("IQR") is defined by $IQR=Q_3-Q_1$. Therefore we compute it using our information from part (a):
$$IQR = Q_3-Q_1 = 47-25=22.$$ Finally, for (c), outliers are any data points lying below $Q_1-1.5IQR$ or above $Q_3+1.5IQR$. First compute $$1.5IQR=1.5(22)=33.$$ Are there any data points below the following value?:
$$Q_1-1.5IQR=25-33=-8$$ The answer is no. Are there any data poitns above the following value?:
$$Q_3+1.5IQR=47+33=77$$ The answer is no. Therefore there are no outliers for this data.

Section 2.5 #16: Find the "five number summary" of the data represented by the following box-and-whisker plot:

Solution: The "five number summary" is the following:
1. minimum=$100$
2. maximum=$320$
3. $Q_1=130$
4. $Q_2=205$
5. $Q_3=270$

Section 2.5 #47: A certain brand of tire has mean life span of $35,000$ miles, with a standard deviation of $2250$ miles. Assume the life spans of the tires have a bell-shaped distribution.
a.) The life spans of three randomly selected tires is $34,000$ miles, $37,000$ miles, and $30,000$ miles. Find the $z$-score that correponds to each life span. Determine whether any of these life spans are unusual.

Solution: We are told that $\mu=35,000$ and $\sigma=2250$. The $z$-score for the data point $x=34,000$ is $$z=\dfrac{x-\mu}{\sigma} = \dfrac{34,000-35,000}{2250} = -0.444\ldots.$$ The $z$-score for the data point $x=37,000$ is $$z = \dfrac{x-\mu}{\sigma} = \dfrac{37,000-35,000}{2250} = 0.888\ldots,$$ and the $z$-score for the data point $x=30,000$ is $$z = \dfrac{x-\mu}{\sigma} = \dfrac{30,000-35,000}{2250} = -2.222\ldots.$$ Since the first two data points' $z$-score is in the interval $(-2,2)$, they are not considered unusual. However, the third data point has a $z$-score less than $-2$, and so it is considered unusual.

Section 3.1 #19: Identify the sample space of the probability experiment and determine the number of outcomes in the smaple space:
determining a person's blood type (A, B, AB, O) and Rh-factor (positive, negative)
Solution: The sample space is comprised of combinations of blood types and Rh-factors, which we list: $$\{A+,A-,B+,B-,AB+,AB-,O+,O-\}.$$ This means that there are $8$ outcomes in the sample space.

Section 3.1 #22: Determine the number of outcomes in the event. Decide whether the event is simple or not: a computer is used to randomly selesct a number from $1$ to $4000$. Event $B$ is selecting a number less than $500$.
Solution: The number of outcomes in Event $B$ is 499. There are that many numbers bigger than or equal to $1$ but also ("strictly") less than 500 (we do not count 500 itself because 500 is not "less than 500").

Section 3.1 #27: A realtor uses a lock box to store the keys to a house that is for sale. The access code for the lock box consists of four digits. The first digit cannot be zero and the last digit must be even. How many different codes are available?
Solution: Recall that "digits" refer to the symbols "0", "1", "2", "3", "4", "5", "6", "7", "8", and "9". There are a total of ten digits. The problem told us that we cannot have the first digit of the code be zero -- this means there are 9 total choices for the first digit of the lock box code. For the second and third digits of the lock box code, we are not told a restriction, and so there are 10 choices for each. For the fourth digit of the code, we may pick any odd digit: there are 5 total choices here. This means there are a total of $$9 \cdot 10 \cdot 10 \cdot 5=4500$$ choices for a code.