Math 251, Final Review Questions

Math 251, Final Review Questions

Name: ANSWERS (to questions 5,6,9 and 15).

5. Consider the data (which are systolic blood pressures of 25 subjects):

105108110110112112116118 118120

126126128130130130 132134136140

146152166188190

(a) What class width should be chosen if you would like to have 8 classes?

ANS: First, the range divided by the number of classes is (190 – 105)/8 = 10.625. Now go up to the next who number to make sure all data are covered in the 8 classes, so we choose a class width of 11.

(b) Complete the following table for this data given that the first class has limits 105—119

Lower

Limit

/ Upper

Limit

/ Lower Boundary / Upper Boundary /

Midpoint

Frequency

/ Cumulative

Frequency

/ Relative Frequency
105 / 119 / 104.5 / 119.5 / 112 / 9 / 9 / .36
120 / 134 / 119.5 / 134.5 / 127 / 9 / 18 / .36
135 / 149 / 134.5 / 149.5 / 142 / 3 / 21 / .12
150 / 164 / 149.5 / 164.5 / 157 / 1 / 22 / .04
165 / 179 / 164.5 / 179.5 / 172 / 1 / 23 / .04
180 / 194 / 179.5 / 194.5 / 187 / 2 / 25 / .08

(c) Find the median, Q1, and Q3 for the above data. Draw a box and whisker plot for the data. You may draw it horizontally if you prefer.

ANS: The median is in the (25 + 1)/2th place, i.e. the 13th place, so the median is 128.

The first quartile is the median of the first 12 numbers, which is 114 (average of 6th and 7th data).

The third quartile is the median of the highest 12 numbers which is 138 (average of 19th and 20th data).

High: 190

Third Quartile: 138

Median: 128

First Quartile: 114

Low: 105

See text for method of constructing box and whisker plot. The lower whisker starts at 105 and goes to 114, the low edge of the box is at 114, the upper edge is at 138, and the line in the box is at 128. The upper whisker starts at 138 and goes up to 190.

(d) Construct a relative frequency histogram for the data using the table in (b).

6. Consider the data (which are systolic blood pressures of 50 subjects):

100,102,104,108,108,110,110,112,112,112,115,116,116,118,118,

118,118,120,120,126,126,126,128,128,128,130,130,130,130,130,

132,132,134,134,136,136,138,140,140,146,148,152,152,152,156,

160,190,200,208,208

(a) What class width should be chosen if you would like to have 6 classes.

ANS: (208 – 100)/6 = 18. Go to the next higher whole number to ensure that all of the data is covered. Thus a class width of 19 would be suitable.

(b) Suppose you don’t want a class width of 19, but would like a class width of 15 irrespective of how many classes that would give you. Complete the following table for this data.

Lower

Limit

/ Upper

Limit

/ Lower Boundary / Upper Boundary /

Midpoint

Frequency

/ Cumulative

Frequency

/ Relative Frequency
100 / 114 / 99.5 / 114.5 / 107 / 10 / 10 / .20
115 / 129 / 114.5 / 129.5 / 122 / 15 / 25 / .30
130 / 144 / 129.5 / 144.5 / 137 / 14 / 39 / .28
145 / 159 / 144.5 / 159.5 / 152 / 6 / 45 / .12
160 / 174 / 159.5 / 174.5 / 167 / 1 / 46 / .02
175 / 189 / 174.5 / 189.5 / 182 / 0 / 46 / .00
190 / 204 / 189.5 / 204.5 / 197 / 2 / 48 / .04
205 / 219 / 204.5 / 219.5 / 212 / 2 / 50 / .04

(d) Draw a frequency polygon using the table in (b).

(e) Draw an Ogive using the table in (b).

f) Find the median, mode, range and first and third quartiles for the data in this problem.

Range: 108

Third Quartile: 140 (the median of the largest 25 numbers is in the 38th place)

First Quartile: 116 (the median of the first 25 numbers in the 13th place)

Median: 129 (the median is the average of the 25th and 26th places)

Mode: 130 (the most commonly occurring data)

9. (a) Make a stem and leaf display for the following data.

5852688672669789849191

9266688786736170757273

85849057777684935847

ANS: (not exactly a stem and leaf display, but a similar table)

4
5
6
7
8
9 / 7
2 7 8 8
1 6 6 8 8
0 2 2 3 3 5 6 7
4 4 4 5 6 6 7 9
0 1 1 2 3 7 4|7 = 47

(b) After making the display, find the median of the data.

ANS: The median is the average of the 16th and 17th places which is 75.5

15. (a) (1 pt) Fill in the missing probability for the following discrete random variable:

X / 3 / 6 / 9 / 10 / 12
P(x) / .10 / .15 / .25 / ? / .11

ANS:

? = 1 - 0.61 = 0.39

(b) The number of cars per household in a small town is given by

Cars 0123

Households 20 280 75 25

(i) Make a probability distribution for x where x represents the number of cars per household in this small town.

ANS:

Cars (x) / 0 / 1 / 2 / 3
P(x) / .0500 / .7000 / .1875 / .0625

(ii) Find the mean and standard deviation for the random variable in (i)

ANS:

Mean:  = 0(.0500)+1(.7000)+2(.1875)+3(.0625) = 1.2625

Variance: 2 = 1(.7000)+4(.1875)+9(.0625) – 1.26252 = 0.41859375

Standard Deviation:  = (0.41859375)1/2 = 0.64699

(iii) What is the average number of cars per household in that small town? Explain what you mean by average.

ANS:

On average, there are 1.2625 cars per household; this is the mean number of cars per household. This is the number one would get if they took the total number of cars in the town and divided by the total number of households.

(c) Different Random Variable Question: (From p. 219 #15)Combinations of Random Variables. Norb and Gary entered in a local golf tournament. Both have played the local course many times. Their scores are random variables with the following means and standard deviations.

Norb, x1: 1 = 115; 1 = 12 Gary, x2: 2 = 100; 2 = 8

Assume that Norb’s and Gary’s scores vary independently of each other. The difference between their scores is W = x1- x2. Compute the mean, variance and standard deviation for the random variable W.

ANS:

Mean: = 115 – 100 = 15

Variance: 2 = 12122 + 1282 = 144 + 64 = 208.

Standard Deviation:  14.42.

(d) Identify the following random variables as continuous or discrete.

(i)x = the winning time of a horse race

(ii)x = the numbers of customers in Vons at a given day

(iii)x = the number of traffic accidents in LA County in a given year

(iv)x = weight of the winning Jockey at a horse race

(v)x = the length of time it takes a person to drive from Anaheim to Hesperia.

ANS: (i),(iv) and (v) are continuous; (ii) and (iii) are discrete.