Business Statistics ( MTH 105) Assignment #1 Student Name: Student ID: Section: 1 Problem 1: A construction company has submitted bids on two separate state contracts, A and B. The company feels that it has a 60% chance of winning contract A, and a 50% chance of winning contract B. Furthermore, the company believes that it has an 80% chance of winning contract A given that it wins contract B. (a) What is the probability that the company will win both contracts? (b) What is the probability that the company will win contract A or contract B? (c) If the company wins contract B, what is the probability that it will not win contract A? (d) Are the events \"winning contract A\" and \"winning contract B\" independent? Why? 2 Problem 2: A driver encounter two traffic lights in the way to work every morning. Each light either red, green or yellow. The probabilities of various combination of colors as is given in the following table: Second Light Red Yellow Green Red 0.28 0.05 0.15 First light Yellow 0.06 0.03 0.1 Green 0.16 0.02 0.15 a) What is the probability that the first light is green b) Find the probability that the first light is red and the second light is green? c) Find the probability that the first light is red or the second light is red? d) If the first light is red, what is the probability that the second light is red? 3 Problem 3: Listed below are the salaries, in $000, for a sample of 16 chief financial officers in the electronics industry. 516 546 486 548 523 558 566 534 586 a) b) c) d) e) f) g) 529 538 Find the 80th percentile Find the five-number summary Calculate the interquartile range (IQR) Find the lower and upper fences Plot the box-plot for the data set What is the shape of the distribution Repeat (b), (c), and (e) using Megastat 4 523 551 552 574 520 Problem 4: Suppose that we will take a random sample of size n from a population having mean and standard deviation . For each of the following situations, find the mean, and standard deviation of the sampling x distribution of the sample mean : a ) = 15, = 2, n = 16 b) = 2, = 2, n = 4 5 Problem 5: Suppose that we will randomly select a sample of 64 measurements from a population having a mean equal to 25 and a standard deviation equal to 8. (38 points) a)Describe the shape of the sampling distribution of the sample mean x . b) Do we need to make any assumptions about the shape of the population? Why or why not? c) Find the mean and the standard deviation of the sampling x distribution of the sample mean . d) Calculate the probability that we will obtain a sample mean greater than 26. e) Calculate the probability that we will obtain a sample mean less than 23. f) Calculate the probability that we will obtain a sample mean between 23 and 26. 6 Problem 6: In each of the following cases, determine whether the sample size n is large enough to say that the sampling distribution of p is a normal distribution. And if yes find the mean and the standard deviation of the sampling distribution of p. a) p = 0.1, n = 10 b) p = 0.1, n = 60 7 Problem 7: 50% of Bank of America customers are satisfied with the bank services. A random sample of 100 customers is taken a) Find the probability that the sample proportion of satisfied customers would be more than 0 .575 b)Find the probability that the sample proportion of satisfied customers would be less than 0.475 c) Find the probability that the sample proportion of satisfied customers would be between 0.475 and 0.575 8 Business Statistics ( MTH 105) Assignment #1 Student Name: Student ID: Section: 1 Problem 1: A construction company has submitted bids on two separate state contracts, A and B. The company feels that it has a 60% chance of winning contract A, and a 50% chance of winning contract B. Furthermore, the company believes that it has an 80% chance of winning contract A given that it wins contract B. (a) What is the probability that the company will win both contracts? Solution:- P(A)=0.6; P(not A)=0.4 P(B)=0.5; P(not B)=0.5 P(A|B)=0.8; P(not A|B)=0.2 P(A and B)= P(A|B)*P(B)=0.8*0.5=0.4 Comment: Notice that P(A and B) is not equal to P(A)*P(B) because 0.4 is not equal to 0.6*0.5=0.3 Therefore events A and B are dependent. (b) What is the probability that the company will win contract A or contract B? Solution:-P(at least one)= 1 - P(win none)= 1- 0.4*0.5=0.8 (c) If the company wins contract B, what is the probability that it will not win contract A? Solution - P(notA |B)= P(not A and B)/P(B) = 0.4*0.5/0.5 = 0.4 (d) Are the events \"winning contract A\" and \"winning contract B\" independent? Why? Solution :-No, the events \"winning contract A\" and \"winning contract B\" are not independent.. P(A)=0.6,P(B)=0.5,P(A|B)=0.8 P(A|B) is not equal to P(A) 2 Problem 2: A driver encounter two traffic lights in the way to work every morning. Each light either red, green or yellow. The probabilities of various combination of colors as is given in the following table: Second Light Red Yellow Green Red 0.28 0.05 0.15 0.48 First light Yellow 0.06 0.03 0.1 0.19 Green 0.16 0.02 0.15 0.33 0.5 0.5 0.1 0.4 1.00 a) What is the probability that the first light is green Solution :The probability that the first light is green= 0.05/1= 0.05 b) Find the probability that the first light is red and the second light is green? Solution :The probability that the first light is red and the second light is green= 0.28/0.48 + 0.15/0.48 = 0.58 +0.3125 =0.8925 c) Find the probability that the first light is red or the second light is red? Solution :- The probability that the first light is red or the second light is red= 1-0.8925 = 0.1075 d) If the first light is red, what is the probability that the second light is red? Solution :The probability that the second light is red= 0.28/0.50 = 0.56 3 Problem 3: Listed below are the salaries, in $000, for a sample of 16 chief financial officers in the electronics industry. 516 546 486 548 523 558 566 534 586 529 538 523 551 552 574 520 a) Find the 80th percentile Solution :- = 16/(100-80)= 16/20 =0.8 =80% b) Find the five-number summary Solution:- 486,516,520,523,523,529,534,538,546,548,551,552,558,566,574,586 Minimum =486 Maximum = 586 Median= (538+546)/2 = 542 Q1= (516+520+523+523+529+534)/6 =524.17 Q2= (548+551+552+558+566+574)/6 =558.17 c) Calculate the interquartile range (IQR) Solution :- IQR=558.17-524.17=34 d) Find the lower and upper fences. Solution :- LF=475 UF= 603 e) Plot the box-plot for the data set. f) What is the shape of the distribution g) Repeat (b), (c), and (e) using Megastat. Solution = Minimum =486, Maximum = 586 Q1= 524.17 Q2= 558.17 Median= 542 IQR= 34 LF=475 UF= 603 4 Problem 4: Suppose that we will take a random sample of size n from a population having mean and standard deviation . For each of the following situations, find the mean, and standard deviation of the sampling x distribution of the sample mean : a ) = 15, = 2, n = 16 Solution - Mean=15 Standard Deviation =2/sqrt 16 = 2/4 = 1/2 b) = 2, = 2, n = 4 Solution :- Mean=2 Standard Deviation =2/sqrt4 =2/2 =1 Problem 5: Suppose that we will randomly select a sample of 64 measurements from a population having a mean equal to 25 and a standard deviation equal to 8. (38 points) a)Describe the shape of the sampling distribution of the sample mean x . Solution :- Since n = 64, the sampling distribution of x is approximately (close to) normal distribution. As such, we assume that it is indeed normal. b) Do we need to make any assumptions about the shape of the population? Why or why not? Solution :- no, because the sample size is large. c) Find the mean and the standard deviation of the sampling x distribution of the sample mean . Solution :- Mean =25 5 Standard Deviation =8/sqrt 84=1 d) Calculate the probability that we will obtain a sample mean greater than 26. Solution :- z = (x-bar - )/(/n) = (26-25)/1=1/1=1 P-value =0.158655 e) Calculate the probability that we will obtain a sample mean less than 23. Solution :- z = (x-bar - )/(/n) Z=(23-25)/1=-2 P(z<2)=0.02275 f) Calculate the probability that we will obtain a sample mean between 23 and 26. Solution :p=0.15866-0.02275 =0.13611 6 Problem 6: In each of the following cases, determine whether the sample size n is large enough to say that the sampling distribution of p is a normal distribution. And if yes find the mean and the standard deviation of the sampling distribution of p. a) p = 0.1, n = 10 Solution :Mean=0.1 Standard Deviation = sqrt (p(1-p)/n) = sqrt (0.1(1-0.1)/10) = 0.095 b) p = 0.1, n = 60 Solution :Mean= 0.1 Standard Deviation = sqrt (p(1-p)/n) =sqrt (0.1(1-0.1)/60) =0.039 7 Problem 7: 50% of Bank of America customers are satisfied with the bank services. A random sample of 100 customers is taken Solution :- Standard Deviation = sqrt (p(1-p)/n) =sqrt (0.50(1-0.50)/100) =0.05 a) Find the probability that the sample proportion of satisfied customers would be more than 0 .575. Solution :- Z-value z = (x - )/() =(0.575-0.50)/0.05 =0.075/0.05 =1.5 P-value(z>0.575) =0.066807 b)Find the probability that the sample proportion of satisfied customers would be less than 0.475. Solution :(0.475-0.50)/0.05 =-0.05 =0.480061 (c) Find the probability that the sample proportion of satisfied customers would be between 0.475 and 0.575 Solution :- p=0.480061-0.066807 =0.413254 8 \fBest solution.. (a)Null hypothesis.. H0: There is no difference in the mean rate of return among the three types of stock, that is m1 = m2 = m3 (Alternative hypothesis) Ha: At least two of the three stocks have different mean rates of return, that is at least two of m1, m2 and m3 are unequal.. p value = 0.0007, which is less than a = 0.05. Therefore, we reject null hypothesis and accept alternative hypothesis.. Conclusion: At least two of the three stocks have different mean rates of return, that is at least two of m1, m2 and m3 are unequal..... (b) No. The analyst , on rejecting H0 is just condent that two of the three stocks have different mean rates of return, but can not be sure if the rates are different for Utility and Retail (It could be Utility and Banking or Retail and Banking too). (4) (a) Mean=15 Standard Deviation =2/sqrt 16 = 2/4 = 1/2 (b) Mean=2 Standard Deviation =2/sqrt4 =2/2 =1 (5) (3) Since n = 64, the sampling distribution of x is approximately (close to) normal distribution. As such, we assume that it is indeed normal. (b) no, because the sample size is large. (c) Mean =25 Standard Deviation =8/sqrt 84:1 (d) z = (itbar u)/(0Nn) = (26-25]/1=1/1=1 Pwalue =0.15865S (e): = (Jr-bar - u)/(oNn) z=(2325)/1=-2 P(z<2)=0.02275 (f) p=0.15866-0.02275 :0. 13611 (6) (a) Mean=0.1 Standard Deviation = sqrt (ptl-pJ/n) = sqrt (o.1(1a.1)/10) = 0.095 (b) Mean= 0.1 Standard Deviation = sqrt (p(1-p]/n) =sqrt (O.1(1O.1)/60) =0.039 (7) Standard Deviation = sqrt (p(1p}/n) =sqrt (0.50( 143.50)! 100) =0.05 (a) Zvalue z = (x - HMO) =(o.575-0.50)/0.05 =0.075/0.05 =1.5 P~va|ue(z>0.575) =0.066807 (b) (o.475-o.50)/0.o5 =0.05 =0.48006 '.'|. (c) p=0.4800610.066807 =0.413254