The individual score of Statistics & Probability Class was reported as below
72 73 78 78 72 75 70 78 92 82
87 93 88 55 77 90 78 63 73 73
85 65 92 80 62 77 73 73 49 80
a) Organize the data using stem-leaf display (4 points)
b) Calculate the mean and standard deviation (4 points)
c) Create the box plot of the data (show the formula and how the quartiles are calculated ! not only the answers). (4 points)
d) Calculate the skewness of the data using Pearson’s method, and name the shape of the frequency polygons based on the skewness. (4 points)
e) Define the range of values which is categorized into outliers. (4 points)
b) Mean: 76.1 Standard Deviation: 10.32
c) Min: 49 Max:93 Q1:72 Q2:77 Q3:82.75 then draw the box plot using these values
d) skewness= 3(mean-median)/standarddeviation = 3(76.1-77)/10.33 = -0.26 (moderately negative skewed)
e) Outliers > Q3+1.5(Q3-Q1) or < Q1-1.5(Q3-Q1)
Q3+1.5(Q3-Q1) = 98.75 Q1-1.5(Q3-Q1)=55.875
Hence, outliers are defined for values > 98.875 or < 55.875
PROBLEM 2 (Chapter 5, Bayes Theorem)
A cell phone salesperson has kept records on the customers who visited the store. 40% of the customers who visited the store were female. Furthermore, the data show that 35% of the females who visited his store purchased a cell phone, while 20% of the males who visited his store purchased a cell phone. Let A1 represent the event that a customer is a female, A2 represent the event that a customer is a male, and B represent the event that a customer will purchase a phone.
a) What is the probability that a male customer will purchase a cell phone?
Based on above description, since 20% of males who visited the store purchased a phone, the probability that a male customer will purchase a cell phone is P(B|A2)=20% (0.2)
b) What is the probability that a female customer will purchase a cell phone?
Based on above description, since 35% of the females who visited the store purchased a phone, the probability that a female customer will purchase a cell phone is P(B|A1)=35% (0.35)
c) The salesperson has just informed us that a cell phone was purchased. What is the probability that customer was female?
Based on Bayes Theorem,
P(A1|B)= ( P(B|A1)xP(A1) ) / ( P(B|A1)xP(A1) + P(B|A2)xP(A2) ) = (0.35×0.4) / (0.35×0.4 + 0.2×0.6) = 0.14/(0.14+0.12) = 0.14/0.26 = 0.538
Just to remind you:
P(A1|B) is the posterior probability
P(B|A1) is the likelihood, P(1) is the prior probability
The denominator P(B|A1)xP(A1) + P(B|A2)xP(A2) is evidence.
Bayes Theorem defined posterior probability as likelihood x prior probability / evidence
PROBLEM 3 (Chapter 6, Discrete Probability Distributions)
Sponsors of a local charity decided to attract wealthy patrons to its $500-a-plate dinner by allowing each patron to buy a set of 5 tickets for the gaming tables. The chance of winning a prize for each of the 5 plays is 50-50. If you bought 5 tickets, what is the chance of winning 4 or more prizes?
This problem can be solved by using Binomial Distribution formula,
P(x) = C(x,n) x p^x x (1-p)^(n-x)
p=0.5, n=5, x=more than 4 = 4 and 5
P(x=4)=C(4,5) x 0.5^4 x 0.5^1 = 0.15625
P(x=5)=C(5,5) x 0.5^5 x 0.5^0 = 0.03125
P(x more than 4) = 0.15625 + 0.03125 = 0.1875
PROBLEM 4 (Chapter 7, Continuous Probability Distribution)
A large manufacturing firm tests job applicants who recently graduated from college. The test scores are normally distributed with a mean of 500 and a standard deviation of 50. Management is considering placing a new hire in an upper level management position if the person scores in the upper 6 percent of the distribution. What is the lowest score a college graduate must earn to qualify for a responsible position? (20 points)
Upper 6% of the distribution, means that the area between the mean (500) and x is 0.5-0.06 = 0.44. Refer to the Table /1 (Areas under normal curve), the area is 0.44 when z=1.56.
1.56 = (x-500)/50
x=(1.56×50) + 500 = 578
The lowest score is 578.
PROBLEM 5 (Chapter 8, Sampling Limit & Central Limit Theorem)
Human Resource Consulting (HRC) is surveying a sample of 60 firms in order to study health care costs for a client. One of the items being tracked is the annual deductible that employees must pay. The state Bureau of Labor reports the mean of this distribution is $502 with a standard deviation of $100.
Compute the standard error of the sample mean for HRC
What is the chance HRC finds a sample mean between $477 and $527 ?
standard error of the sample mean is obtained by 100/root(60) = 100/7.746=12.91
z1=(477-502)/12.91 = -1.94 (area=0.4738) z2=(527-502)/12.91 = 1.94 (area=0.4738)
The probability is found by 0.4738+0.4738 = 0.9476