Question

Five state officials are listed below: Lieutenant Governor (L) Secretary of state (S) Attorney General (A) Representative (R) Press Secretary (P) a. List all 10 possible samples (without replacement) of size 2. Use the letter abbreviation of each official. b. What is the chance that the first sample in your list is selected? The data show the number of hours of television watched per day by a sample of 11 people: 6 7 1 2 5 6 9 3 3 4 2 a. obtain the quartiles: (type integers or decimals) Q1 = Q2 = Q3 = b. Calculate the interquartile range. IQR = c. Find the five-number summary

Answer 1

First Part. a. The 10 possible samples (without replacement): {SL, AL, RL, PL, AS, RS, PS, RA, PA, PR}; b. The chance that the first sample in the list is selected is 1/10. Second Part. a. Q1 = 2, Q2= 4, Q3 = 6; b. The interquartile range (IQR) = Q3 - Q1 = 6 - 2 = 4; c. Five-number summary: a. 1; b. 2; c. 4; d. 6; e. 9.

Explanation:

First Part

a. List all possible samples (without replacement):

There are five different state officials (L, S, A, R, P).

Samples of size 2 from this set are determined by combinations, since this is a situation where the order in which the objects are taken is meaningless: samples (L, S) = (S, L), or, it is the same having a set of a Lieutenant Governor and a Secretary of State than a Secretary of State and a Lieutenant Governor.

So, the number of possible samples (without replacement) is
`\\ (n!)/((n-k)!k!)`, where n is the number of elements from which we can make different samples of k = 2. Then,
`\\ (5!)/((5-2)!2!) = (5!)/(3!2!) =(5*4*3!)/(3!2!) = (5*4)/(2*1) = 5*2 = 10.`

We can make five different sets as follows considering all the five different elements {L, S, A, R, P}:

1. {LL, SL, AL, RL, PL}
2. {LS, SS, AS, RS, PS}
3. {LA, SA, AA, RA, PA}
4. {LR, SR, AR, RR, PR}
5. {LP, SP, AP, RP, PP}

But we know that samples like {LL, SS, AA, RR, PP} cannot be possible since there must be one L, one, S and so on. Thus, for the 25 case we have to subtract 5, leaving 20 samples.

Likewise, we know, for the reason aforementioned, that samples SL and LS are the same, as AL = LA, AS = SA, and the like. Because for each sample there is a duplicated, we have to divide the 20 samples by 2 (or 2!), so the samples are, therefore, 20/2 = 10, or:

1. {SL, AL, RL, PL}
2. {AS, RS, PS}
3. {RA, PA}
4. {PR}

Then, the 10 possible samples (without replacement) are: {SL, AL, RL, PL, AS, RS, PS, RA, PA, PR}.

b. The chance to select the first sample of the list is 1/10 or 10% of chance to obtain the first sample {SL}, since we have 10 different cases, and only a case for each sample.

Second Part

a. Obtain the quartiles for {6, 7, 1, 2, 5, 6, 9, 3, 3, 4, 2}.

Quartiles are percentiles 25th, 50th and 75th, or the number from a set of data below which are 25%, 50% and 75% of values for this set of data.

Firstly, we have to order the set from minimum to maximum values, as follows:

{1, 2, 2, 3, 3, 4, 5, 6, 6, 7, 9}

Different methods to find quartiles

Tukey's Method

There are different methods to obtain quartiles from a dataset, and values for them can, therefore, be different as a result.

If we use the Tukey's method to solve find the quartiles, we can proceed as follows:

Find the median of the dataset (or 50th percentile, or second quartile): we have an odd number of values (n = 11), then, what is the value in the dataset that have 50% of values below it and 50% of values above it? This value is 4, because there are five numbers below it {1, 2, 2, 3, 3}, and five numbers above it {5, 6, 6, 7, 9}. Then, 4 is the median, 50th percentile, or second quartile (Q2).

To find the 25th percentile (or Q1) and the 75th percentile (or Q3), we can proceed as follows:

Add the median obtained in the previous step to the two halves {1, 2, 2, 3, 3} and {5, 6, 6, 7, 9}; so,

{1, 2, 2, 3, 3, 4} and {4, 5, 6, 6, 7, 9}.

To find Q1 and Q3 find the medians of each dataset as previously described.

Then,

Median of {1, 2, 2, 3, 3, 4} is (2+3)/2 = 2.5, since we have even number of elements. So, Q1 = 2.5.

Median of {4, 5, 6, 6, 7, 9} is (6+6)/2 = 6 (the same rule as before). So, Q3 = 6.

TI-83 (described by Moore and McCabe)

Similar to that of Tukey's but not including the median. Then,

Median is also 4 for the same reasons given before. The two halves are:

{1, 2, 2, 3, 3} and {5, 6, 6, 7, 9}

The median for each dataset:

Median of {1, 2, 2, 3, 3} = 2 = Q1.

Median of {5, 6, 6, 7, 9} = 6 = Q3.

There are many more methods and obviously different results.

b. The interquartile range is Q3 - Q1. Considering the last result for Q1, Q2 and Q3, then:

The interquartile range is Q3 - Q1 = 6 - 2 = 4.

c. The five-number summary is:

• The sample minimum: in this case, for all values, 1 is the minimum of all of the dataset.
• The first quartile (Q1) = 2.
• The second quartile (Q2) = 4.
• The third quartile (Q3) = 6.
• The sample maximum: in this case the maximum value of the dataset is 9.