QMTH 205 Business Statistics Problem Set 2

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Problem Set 2 QMTH 205 Spring 2019 1. The following table shows the outcome of 500 interviews attempted during a survey of options about big business held by residents in a Rock Hill. The data are also classified by the area of the city in which the interview was attempted. A questionnaire was selected at random from the 500. Area of City A B C D Total a. b. c. d. e. f. g. Completed 100 115 50 35 300 Not at Home 20 5 60 50 135 Refused 5 5 15 40 65 Total 125 125 125 125 500 What is the probability that the questionnaire was completed? What is the probability that the potential respondent was not at home? What is the probability that the potential respondent refused to answer? What is the probability that the potential respondent lived in Area B? What is the probability that the potential respondent lived in Area A and refused to answer? What is the probability that the potential respondent was not at home or refused to answer? What is the probability that the potential respondent was from Area D or completed the questionnaire? h. What is the probability that the potential respondent was from Area D and completed the survey? i. What is the probability that the potential respondent was not at home, given that he was from Area D? j. What is the probability that the potential respondent was from Area B given that he refused to answer? 2. In a large city, 70% of the households have cable Internet access and 90% have a smart tv. Suppose these two events are independent. What is the probability that a randomly selected household will be one that has cable access and has a smart tv? 3. In a certain firm, 60% of the employees are males. Furthermore, 80% of the females and 60% of the males are high school graduates. Find the probability that an employee picked at random will be: a. a male high school graduate. b. A female high school graduate Problem Set 2 QMTH 205 Spring 2019 4. The personnel director of a firm has determined the following probabilities for the length of time a certain type of employee remains with the firm after being hired. Length of Stay Probability Less than 1 year 0.05 1 year but less than 2 0.10 2 years but less than 3 0.10 3 years but less than 4 0.15 4 years but less than 5 0.20 5 years or longer 0.40 Total 1.00 Find the probability that a newly hired employee of this type will be with the firm: a. Less than 5 years b. 3 years or more c. At least 2 but less than 4 years d. At least 4 years 5. With a single roll of a pair of dice, determine the probability of getting the following: a. The sum of 7 b. The sum of 7 given that at least one of the dice is a 5 c. A 4, 5, or 6 on at least one of the dice d. A sum that is odd or more than 5 e. A sum that is odd and more than 5

Tutor Answer

apradeep92
School: Cornell University

Please find all solutions in attached doc and PDF. Please note both files are exactly same content-wise. Please let me know, if you have any doubt.Thanks

1. Solutions:
a. π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (π‘žπ‘’π‘’π‘ π‘‘π‘–π‘œπ‘›π‘›π‘Žπ‘–π‘Ÿπ‘’ π‘€π‘Žπ‘  π‘π‘œπ‘šπ‘π‘™π‘’π‘‘π‘’π‘‘) =

π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘œπ‘šπ‘π‘™π‘’π‘‘π‘’π‘‘ questionnaire
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ questionnaires

Using given table:
π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘œπ‘šπ‘π‘™π‘’π‘‘π‘’π‘‘ questionnaire = 300
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ questionnaires = 500
π‘ π‘œ, π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (π‘žπ‘’π‘’π‘ π‘‘π‘–π‘œπ‘›π‘›π‘Žπ‘–π‘Ÿπ‘’ π‘€π‘Žπ‘  π‘π‘œπ‘šπ‘π‘™π‘’π‘‘π‘’π‘‘) =

300
= 𝟎. πŸ”
500

b. π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦(π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ π‘€π‘Žπ‘  π‘›π‘œπ‘‘ π‘Žπ‘‘ β„Žπ‘œπ‘šπ‘’) =

π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ π‘›π‘œπ‘‘ π‘Žπ‘‘ β„Žπ‘œπ‘šπ‘’
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ questionnaires

Using given table:
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ π‘›π‘œπ‘‘ π‘Žπ‘‘ β„Žπ‘œπ‘šπ‘’ = 135
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ questionnaires = 500
π‘ π‘œ, π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ π‘€π‘Žπ‘  π‘›π‘œπ‘‘ π‘Žπ‘‘ β„Žπ‘œπ‘šπ‘’) =

c. 𝑃(potential respondent refused to answer) =

135
= 𝟎. πŸπŸ•
500

π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ refused to answer
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ questionnaires

Using given table:
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ refused to answer = 65
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ questionnaires = 500
π‘ π‘œ, π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ refused to answer) =
d. 𝑃(potential respondent lived in Area B ) =

65
= 𝟎. πŸπŸ‘
500

π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ in Area B
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ questionnaires

Using given table:
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ in Area B = 125
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ questionnaires = 500
π‘ π‘œ, π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ lived in Area B ) =

125
= 𝟎. πŸπŸ“
500

e. 𝑃(potential respondent lived in Area A and refused to answer ) =
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ in Area A, who refused to answer
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ questionnaires
Using given table:
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ in Area A, who refused to answer = 5
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ questionnaires = 500
5
π‘ π‘œ, π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ π‘€π‘Žπ‘  π‘›π‘œπ‘‘ π‘Žπ‘‘ β„Žπ‘œπ‘šπ‘’) =
= 𝟎. 𝟎𝟏
500

f.

𝑃(potential respondent was not at home or refused to answer ) =
𝑃(potential respondent was not at home) + 𝑃(potential respondent refused to answer)
βˆ’ 𝑃(potential respondent was not at home and refused to answer )

πΉπ‘Ÿπ‘œπ‘š π‘π‘Žπ‘Ÿπ‘‘ 𝑏:
𝑃(potential respondent was not at home) = 0.27
πΉπ‘Ÿπ‘œπ‘š π‘π‘Žπ‘Ÿπ‘‘ 𝑐:
𝑃(potential respondent refused to answer) = 0.13
As, Potential respondent not being at home and Potential respondent refusing to answer
are mutually exclusive events, so
𝑃(potential respondent was not at home and refused to answer ) = 0
So,
𝑃(potential respondent was not at home or refused to answer) = 0.27 + 0.13 βˆ’ 0
= 𝟎. πŸ’πŸŽ
g. 𝑃(potential respondent was from Area D or completed the questionnaire ) =
𝑃(potential respondent was from Area D )
+ 𝑃(potential respondent completed the questionnaire)
βˆ’ 𝑃(potential respondent wasfrom Area D and completed the questionnaire )

𝑃(potential respondent was from Area D )
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ in Area D
=
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ questionnaires
Using given table:
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ in Area D = 125
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ questionnaires = 500
125
π‘ π‘œ, π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ lived in Area D ) =
= 0.25
500
πΉπ‘Ÿπ‘œπ‘š π‘π‘Žπ‘Ÿπ‘‘ π‘Ž:
𝑃(potential respondent completed the questionnaire) = 0.6
𝑃(potential respondent was from Area D and completed the questionnaire )
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ in Area D, who completed the questionnaire
=
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ questionnaires
Using given table:
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ in Area D, who completed the questionnaire = 35
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ questionnaires = 500
35
π‘ π‘œ, π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ lived in Area D ) =
= 0.07
500
So,
𝑃(potential respondent was from Area D or completed the questionnaire )
= 0.25 + 0.6 βˆ’ 0.07 = 0.85 βˆ’ 0.07 = 𝟎. πŸ•πŸ–

h. 𝑃(potential respondent was from Area D and completed the questionnaire ) =
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ in Area D, who completed the questionnaire
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ questionnaires
Using given table:
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ in Area D, who completed the questionnaire = 35
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ questionnaires = 500
35
π‘ π‘œ, π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ (π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ lived in Area D ) =
= 𝟎. πŸŽπŸ•
500

i.

𝑃(potential respondent was not at home, given that he was from Area D )
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ in Area D, who were not at home
=
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ in Area D
Using given table:
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘Ÿπ‘’π‘ π‘π‘œπ‘›π‘‘π‘’π‘›π‘‘ in Area D, who were not at home = 50
π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘œπ‘‘π‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘Ÿπ‘’π‘ π...

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