# 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 27-year-old 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 left-skewed.

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

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 = 13C3

Number of ways of selecting 3 cards out of 52 = 52C3

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 left-skewed 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 = ?x2*P(x) – Mean2 = 26.4864 – 4.962 = 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.

##### No Need To Pay Extra
• Turnitin Report

\$10.00
• Proofreading and Editing

\$9.00
Per Page
• Consultation with Expert

\$35.00
Per Hour
• Live Session 1-on-1

\$40.00
Per 30 min.
• Quality Check

\$25.00

### Free

New Special Offer