Stat 200 Week 5
1.) A doctor wishes to estimate the mean triglyceride level of patients two hours after eating a steak dinner. To estimate the mean, you collect data from 22 patients. State the individual, variable, population, sample, parameter, and statistic.
Individual: Patients
Variable: triglyceride level of patients two hours after eating a steak dinner
Population: All the patients
Sample: 22 patients selected for analysis
Parameter: mean triglyceride level of all the patient’s two hours after eating a steak dinner.
Statistics: mean triglyceride level of 22 patients selected for analysis, two hours after eating a steak dinner.
2.) Explain why the relative frequency column of a frequency table might not sum to 1.0.
It is possible that due to rounding off decimals up to 2 digit or 3 digits after decimal, we can get the sum of all the relative frequencies slightly different from 1 like 1.01 or 0.99. It only occurs due to rounding off error.
3.) The Affordable Care Act created a market place for individuals to purchase health care plans. In 2014, the premiums for a 27yearold for the bronze level health insurance are given in the table below. Create a frequency distribution, relative frequency distribution, and cumulative frequency distribution using 6 classes. (All prices in the table are in dollars $)
201 
189 
203 
193 
179 
181 
178 
164 
177 
190 
189 
205 
199 
174 
195 
191 
180 
197 
156 
203 
175 
202 
170 
179 
200 
202 
204 
184 
203 
199 
188 
198 
194 
208 
206 
205 
Minimum value = 156
Maximum value = 208
Range = 208 – 156 = 52
Number of classes = 6
Class width = 52/6 = 9 (approximately after rounding to next integer)
Class Interval 
Frequency 
Relative Frequency 
Cumulative Frequency 
156  164 
2 
0.056 
2 
165  173 
1 
0.028 
3 
174  182 
8 
0.222 
11 
183  191 
6 
0.167 
17 
192  200 
8 
0.222 
25 
201  209 
11 
0.306 
36 
Total 
36 
1.001 

4.) Students in Stat 200 took their first exam. The scores are located in the table below. Create a frequency distribution and histogram for the data using class limits that make sense for the data. Describe the shape of the distribution.
99 
86 
93 
84 
86 
80 
84 
93 
82 
75 
73 
92 
100 
90 
97 
100 
71 
60 
77 
85 
Minimum value = 60
Maximum Value = 100
Range = 40
I am using a number of classes as 6.
Class width = 40/6 = 7 (approximately after rounding to next integer)
Class Interval 
Frequency 
60  66 
1 
67  73 
2 
74  80 
3 
81  87 
6 
88  94 
4 
95  101 
4 
The shape of the distribution is leftskewed.
5.) Find the mean, median, range, variance and standard deviation of the data set below. The data represent the fixed monthly costs for Sam's Linen Service.
Monthly charges 
Monthly cost ($) 
Bank charges 
482 
Cleaning 
2208 
Food 
1750 
Computer expenses 
2471 
Lease payments 
2656 
Grounds fees 
1475 
Postage 
2117 
Uniforms 
2600 
Delivery fuel 
955 
Mean = (total sum)/n = 16714/9 = 1857.111
Increasing order of [removed]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" style="height:19px; width:697px" />
Variance = 577481.446
Standard deviation = 577481.446 = 759.922
6.) You pick 3 cards from a deck without replacing a card before picking the next card. What is the probability that all 3 cards are hearts?
Number of ways of selecting 3 hearts out of 13 = ^{13}C_{3}
Number of ways of selecting 3 cards out of 52 = ^{52}C_{3}
P(all 3 cards are heart) = 313C352C = 286/22100 = 11/850 =0.0129
7.) The 2010 U.S. Census found the chance of a household having a certain number of pets. The data is in the table below.
# of pets 
0 
1 
2 
3 
4 
5 or more 
Probability 
15.5% 
22.8% 
26.4% 
18.5% 
10% 
6.8% 
Is it unusual for a household to have 4 pets? Show all work.
P(4 pets) = 0.10
As the probability of having 4 pets is more than 0.05, this event is not unusual.
8.) Suppose a random variable, x, arises from a binomial experiment. Suppose n = 8 and p = 0.62.
a.) Write the probability distribution.
b.) Draw a histogram.
c.) Describe the shape of the histogram.
d.) Find the mean.
e.) Find the variance.
f.) Find the standard deviation.
(a) Probability distribution:
x 
P(x) 
0 
0.000435 
1 
0.005675 
2 
0.032407 
3 
0.10575 
4 
0.215675 
5 
0.281512 
6 
0.229655 
7 
0.107057 
8 
0.021834 
(b) Histogram:
(c) Shape of distribution is leftskewed or negatively skewed.
(d) Calculation for mean:
x 
P(x) 
x*P(x) 
0 
0.000435 
0 
1 
0.005675 
0.005675 
2 
0.032407 
0.064815 
3 
0.10575 
0.317251 
4 
0.215675 
0.862699 
5 
0.281512 
1.407561 
6 
0.229655 
1.377928 
7 
0.107057 
0.7494 
8 
0.021834 
0.174672 
Sum 
1 
4.96 
Mean = ?x*p(x) =4.96
(e) Calculation for variance:
x 
P(x) 
x*P(x) 
x^2*P(x) 
0 
0.000435 
0 
0 
1 
0.005675 
0.005675 
0.005675 
2 
0.032407 
0.064815 
0.129629 
3 
0.10575 
0.317251 
0.951752 
4 
0.215675 
0.862699 
3.450795 
5 
0.281512 
1.407561 
7.037806 
6 
0.229655 
1.377928 
8.26757 
7 
0.107057 
0.7494 
5.245797 
8 
0.021834 
0.174672 
1.397377 
Sum 
1 
4.96 
26.4864 
Variance = ?x^{2}*P(x) – Mean^{2} = 26.4864 – 4.96^{2} = 1.8848
(f) Standard deviation = 1.8848 =1.3729
9.) The mean starting salary for teachers is $53,475 nationally. The standard deviation is approximately $5,250. Assume that the starting salary is normally distributed.
a.) State the random variable.
Starting salary of teachers
b.) Find the probability that a starting teacher will make more than $60,000.
Z score for 60,000:
Z = (60000 – 53475)/5250 = 1.24
P(x > 60000) = P(z > 1.24) = 1 – 0.8925 = 0.1075
c.) Find the probability that a starting teacher will make less than $50,000.
Z score for 50,000:
Z = (50000 – 53475)/5250 = 0.66
P(x < 50000 xss=removed>
d.) Find the probability that a starting nurse will make between $52,000 and $59,500.
Z score for 52,000:
Z = (52000 – 53475)/5250 = 0.28
Z score for 59,500:
Z = (59500 – 53475)/5250 = 1.15
P(52000 < x xss=removed>
e.) If a teacher made more than $61,000, would you think the teacher was overpaid? Why or why not?
Z score for 61,000:
Z = (61000 – 53475)/5250 = 1.43
P(z > 1.43) = 0.5 – 0.4236 = 0.0764
As P > 0.05, a teacher making more than 61,000 is not unusual. So, I don’t think the teacher was overpaid.
10.) The following table contains the cholesterol levels of 32 people. Determine if the sample comes from a population that is normally distributed. Draw a histogram, identify potential outliers, and draw the normality plot.
160 
200 
172 
196 
170 
181 
190 
185 
210 
180 
162 
184 
185 
212 
186 
182 
260 
175 
203 
181 
190 
203 
184 
193 
199 
160 
210 
204 
198 
188 
201 
190 
It looks like an outlier in the class interval 244 to 265. The outlier is 260.
Due to the presence of one outlier, it is not a normal distribution. If we remove the outlier then we can get a normal distribution.