Question

b. In the area to the right, create a probability distribution table showing the possible values of X', the frequency of each value, and the associated relative frequency values P(X) as determined by the c. Determine the expected value (mean) of the random variable X using your probability distribution table created in part b. directly above. (Hint: the requirement is to use only the information in the table you produced in part b. not to use the raw data---see how to produce the mean from a probability distribution table via the Excel Guides for Unit 2 or through the equivalent text's method explained in Section 7-2.) MEANU: d. Determine the standard deviation of the random variable x again using only the values within your probability distribution table. (You can check your answers by finding the population s.d of the data on family size, BUT this problem needs to be answered through use of only the probability distribution table built in Dart b..see Section 7-2.1 St. Daviation : 4. Typically, any data outside of the 2 Sigma Rule (pg. 154) is considered "unusual." Decide if any of the included values of the random variable x' are unusual. Give a concluding statement below in regard to your decision. Show work for 2 Sigma Rule to the right of the problem. Create a probability statement which is supported by the values in the probability distribution table. For example, "The probability that a randomly selected member of this group comes from a family of size 7 or

Number in Family
4
4
5
2
8
3
4
6
1
3
7
4
4
3
5
2
4
5
4
3
3
3
6
6
2
6
7
3
2
3
8
5
5
4
3
4
5
4
5
4
8
5
4
4
6
5
6
4
5
2
3
3
6
5
7
3
7
4
2
4
7
8
4
6
4
2
6
3
2
7
4
6
5
4
4
4
3
5
4
8
3
3
8
4
5
6
5
3

Answer 1

b. Probability distribution

Arrange the data in ascending order and count the number of members in each group. Put them in the table as shown below to compute the relative frequency.

Members Frequency Relative Frequency 1 1/88 0.0114 8 8/88 0.0909 17 17/88 0.1932 24 24/88 0.2727 15 15/88 | 0.1705 11 11/88 0.1250 6/88 | 0.0682 6 6/88 0.0682 Total 88

c. Mean and standard deviation

P(2) P(0) () (1 - 1) (x - 72 P(x) x (1 – 7)2 0.0114 0.0909 0.1932 0.2727 0.1705 0.1250 0.0682 0.0682 0.01 0.18 0.58 1.09 0.85 0.75 0.48 0.55 -3.49 -2.49 -1.49 -0.49 0.51 1.51 2.51 3.51 12.1801 6.2001 2.2201 0.2401 0.2601 2.2801 6.3001 12.3201 0.1384 0.5636 0.4289 0.0655 0.0443 0.2850 0.4296 0.8400 L / / 6 7 8 ī= P(x) < (2) 4.49 o? = P(x) x (x – ī)2 2.7953 O=Vo2 = 1.6719

d. Outliers.

mean std. Dev 4.4886 1.6719 Step 2: Find the Lower limit and upper limit for each std dev. Lowerlimit 1 mean - 1 SD 4.4886-(1*1.6719) 2 mean - 2 SD 4.4886-(2*1.6719) 3 mean - 3 SD 4.4886-(3*1.6719) 2.8167|mean + 1 SD 1.1448 mean + 2 SD -0.5271 mean + 3 SD Upper limit 4.4886+(1*1.6719) 4.4886+(2*1.6719) 4.4886+(3*1.6719) 1 6.1605 7.8324 9.5043 Therefore any value lower than 1.14 and greater than 7.8 is an outlier. We have one observation with value = 1 and six observation with value = 8 Hence these observation are considered to be outlier.