Debbie and Katie are planning a visit to the Elegant Day spa where they will eac
ID: 3324960 • Letter: D
Question
Debbie and Katie are planning a visit to the Elegant Day spa where they will each choose one of the following three services that offered. Service Facial (F) Pedicure (P) Manicure () Cost $50 $40 $30 a) Identify the frequency distribution for all combination of sample means for Debbie and Katie choosing a service. b) which Theorem (or Theory) will we assume for the sample mean? 1) Queuing Theory 2) Decision Theory 3) Central Tendency Theorem 4) central Limit Theorem c) Calculate thetandard deviat!ion of the sample mean.Explanation / Answer
(A)
The all combination of sample means for Debbie and Katie choosing a service is (F, F) (F, P) (F, M) (M, F) (M, P) (M, M) (P, F) (P, M) (P, P).
Total option n = 9
We know that, probability P = m/n
where, m is frequency of favourable options
Thus, the frequency distribution for all combination of sample means for Debbie and Katie choosing a service as below :
Services
Average Cost
Frequency m
Probability Distribution P= m/n
(M, M)
30
1
0.1111
(P, M) , (M, P)
35
2
0.2222
(P, P), (F, M) , (M, F)
40
3
0.3333
(F, P) , (P, F)
45
2
0.2222
(F, F)
50
1
0.1111
(B)
We will assume for the sample mean : 4) Central Limit Theorem.
Because, probability distribution is define by Central Limit Theorem.
(C)
Services
Average Cost X
Frequency f
Probability Distribution
X-X'
f(X-X')
f(X-X')^2
(M, M)
30
1
0.1111
-10
-10
100
(P, M) , (M, P)
35
2
0.2222
35
70
4900
(P, P), (F, M) , (M, F)
40
3
0.3333
40
120
14400
(F, P) , (P, F)
45
2
0.2222
45
90
8100
(F, F)
50
1
0.1111
50
50
2500
Average Mean X' =
40
N=9
f(X-X')^2
30000
Standard Deviation= sqrt[{f(X-X')^2}/N] = 57.735
Services
Average Cost
Frequency m
Probability Distribution P= m/n
(M, M)
30
1
0.1111
(P, M) , (M, P)
35
2
0.2222
(P, P), (F, M) , (M, F)
40
3
0.3333
(F, P) , (P, F)
45
2
0.2222
(F, F)
50
1
0.1111
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.