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

Increasing order of [removed]iVBORw0KGgoAAAANSUhEUgAAArkAAAATCAIAAAAbJIJCAAAAAXNSR0IArs4c6QAAAAlwSFlzAAAOxAAADsQBlSsOGwAABsJJREFUeF7tXT12qzoQlu/bQdYALnzeAnJyV8Bx4yptulC6SneLFOloTJnyndfdigazhvQ+FIY1ZAu+MwLDCPMnrsHCZ9REMdJovu/TjGRh48XpdBJcmAFmgBlgBpgBZoAZaGDgBzPDDDADzAAzwAwwA8xACwP/vL+/K5fTNH14eICX0ij6She2DfU0jb7O9T5sFjaIEWkuTffp4nHxjeawLBa2HIuUsi+Om/uC7pB61hwsfi/2/389Ptp9nJJdEEg2KPHH/i59Q7jdxViAVaIyRzP9SsIp9nHx9ldzmKAlXgXj91cx2UBQhFsz05pkNlbcShjWc1vVesTJPLa4FK+CnQpK0lQ3VJpGyo7mpKY832apqi5yyyt9wCrZQBPvlOLmqfn7QS42JFrvNHJxHUKwdXk4v9RL3sGZapi4+bkCeJ2ie5H/8hJCLXLdxHEcO/zpp5EfCqgeX9yoGwG09t0XL5HGiJE0DI742hLW9SSU1WMQJBa1SPs21M9+YjfLcUSyXDuqUzBonZvYMXI9BJV40IL6Q+udCE0GSH2jalYInwavpppDBKVDUIyKuL4b2jCRUfTOYrK4NAybuAWATcFVi30w3gnEpXgV7ERQmmHuRty2yCXZdTy8E4sL2wPf9RPhwGIAtSJaaX08sLifIqyOsu4oCygBS9ZECbC8NB5eTbA0LeNeIfJxYwBKRf5x+fwvOi0Xcyzwn7PdKkt6Kw4LWm/QRMWIEKulY8NKbVloECrwykZd50lf2AgUdmgd9geJ78ttDTLrHY9htslpLRnAJDis8ATCXokj9Cn9UesdtgwGaFHSiJqXhFPsY+HVVHOAoApedVIVANPwt1hawlpnoo8FVoZU7Yy91uxF80UYNo0FbjQFVy3ywZNZE+wQcSleUqeCRjRNdWlLhVDymwGpSRG3JXLVAB9rMk8sbup74m0rz4YVce8ycglYwKvkYfWSgeLCXiEKxBsu2rBTWOeSifVGBH7ke/FqLRMU4vjvs1zaYXfikpKv3io+ixqx1kuRCt+Fc4qsFYxaPRLojHbZwFkLT75FtLafn1vYduQZNHfp43D4QM+ISzlA5+059nzfD2PsTPy58A0BzxCgwh9RM2O8JNwkvIWaAwS9nC85xjqA4hDLwy45m+cqbjUMmyKmEly3wvvX4lK8F9gP8UWaugtxWyK3Tu+5iwv7g4NIUpHA2lBEKERrsQ+8p8ilYFM1TSmXbp2p6iP3tH993cOXIU5Jsoeye3163eGfXYKv7Z7w4n4n/+tbcouJakT23r8+5YazQS/K2Zuscdmoqd5qILtIu0pMitnMH+pbN0yTAZ59I2pmSl4QftZiRLx9FKzIoyso6V7FKAFWtBoRbM8Z24r38iJ5RQ3D5oio0boF9uDJPLq4FG9ZJ/5W0lS3tudcYGBqKgDSPFwXuV3xorAwA3HzNSbL09Tfe4xcBay67tRdap/RU4tbfg8CTnShLGFHs3Sc9Sr2ojQN4+eNE7kfv2PPdX/2ueOLb9qCwyGAznDuS4zALSn42EYgnvGgIvIEvf+A92nkm9+ib0u93+lDfSu8WQMHXnBAAu8sC39ovYd1swGeySdqVgkfGe9kalYmSTmpKEBns4LJGIVi0+tOmrniqmHYFCl4gKQGV/uM1sU7nbgUL60TQZU0NevIJQBpHq5Jlefs2gOudiq+ibjW9pfA5cIP4J40jdZ7jFwFrJqHlUs91L1B5OI7Tq1Tgx6793Gb6Dqs235c73tY13VYt30PF6ZrMsD5AV2mw9M1kq7zuu27xp/0+gDnB3SZFFLrYLrO67Y3B2l2VKm7cAzoYg5kXed125uDtEHcH3Dbf5N/AKDHZsaAJpEX93uLePaVARqgWuOddV015eca5jVjKfa7n71/BZbFNThUK65pz2QWd+7iyt1MovmJhFttgdDPvc5HJ86OMsBbSdYy7mA1ZzRjKfzBeOcye68ClsU1MFQrLg2eySzujMVdgO/z2e6wp8wAM8AMMAPMADPADDADzAAzwAwwA8wAM2ASA3yuYJIa7AszMCED8MWkJLFtfD5ad4Hnwlq9Gnab4hbMADMwOwb4t6NmJxk7zAxcgwH6yPN2e9Vnh19jdLbBDDADs2KA9wqzkoudZQauxAB8UZ8+8rzNqtajha/kHpthBpgBoxjgexBGycHOMANTMYC/pyJWKxHH8HSy/EFVcILgBfgM9Kysigv4Q3DikzzlfSoveRxmgBkwggHeKxghAzvBDNyGAdwxLHtsAnivcBt9eFRmwBAG+B6EIUKwG8zA1AyUjzzvGpk+UrqrLV9nBpiBO2TgD14yD8iutB6XAAAAAElFTkSuQmCC" 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 = 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.