Decision making

ASSIGNMENT ON DECISION SCIENCE FUNDAMENTALS 

SUBMITTED TO:  PROF. MANORANJAN NAYAK 

 

SUBMITTED BY:  Kiran kumar jena

1.What are the three common measures of central tendency? 

Ans. Three common measures of central tendency are the arithmetic mean(mean), median, and mode. 

2. Write the symbol of population mean. 

Ans: The symbol of population mean is µ. 

3. How is median calculated in case of even data? 

Ans. The median calculated in case of even data the median is the mean 0f the two middle observations in the ordered list. 

4.Which measure of central tendency do you use for qualitative data set? 

Ans: Mode is use for qualitative data set. 

5.Which measure of central tendency do you use for quantitative data set? 

Ans. Mean or median is used for quantitative data set. 

6.What do you mean by in outlier? 

Ans: An outlier is a data point that differs significantly from other observation. An outlier may be due to variability measurement or it may indicate experimental error; the latter are sometimes excluded from the data set. An outlier can cause serious problem in the statistical analyses. 

7.What are the three common measures of variability? 

Ans. Three common measures of variability are the range, variance and standard deviation. 

8.How is inter-quartile range calculated? 

Ans: The inter-quartile range calculate are as follows, 

a. Arrange the observations in increasing order and locate the median in the ordered list of observations. 

b. The first quartile is the median of the observations whose position in the ordered list is to the left of the location of the overall median. 

c. The third quartile is the median of the observation whose position in the ordered list is to the right of the location of the overall median. 

9.Write the formula of standard deviation. 

Ans: The formula of standard deviation is  

Standard deviation   

10.What are the significance of variance? 

Ans: The variance is a numerical value used to indicate how to wildly individuals in a group vary. If individual observation difference greatly from the group mean, the variance is big and vice versa. In short, variance measures how far a data set is spread out.  

1.The set S= {5,10,15,20,30}, find mean of set S. 

Ans: 

 Sum of all number÷N=mean

5+10+15+20+30=80

80÷5=16

 

 

X 

 

N=5 

Mean= ∑ 𝑁𝑋 

           =              =16 

Mean= 16 

2.Find the median of 11,22,33,55,66,99. 

Ans: n=6 

Median= {𝑛2 +(𝑛2+ 1)} th item/2 

             = { +( +1)} th item/2 

       = (3+4) th item/2 

       =(33+55)/2 

       =88/2 

       =44 

Median =44 

3.Which of the following sequences have the highest S.D? 

2,4,6,8 

3,6,9,12 

4,8,12,16 

1,2,3,4 

Ans.4,8,12,16 

4.Find the mean, median, mode and range for the following list of values:13,18,13,14,13,16,14,21,13 

Ans. Arrange In ascending order 13,13,13,13,14,14,16,18,21 

x f (frequency) fX CF 

13 4 52 4 

14 2 28 6 

16 1 16 7 

18 1 18 8 

21 1 21 9 

  ∑𝑓 = 9 ∑𝑓𝑥 =135  

 

Mean= ∑ 𝑓𝑓𝑥            =   

           = 15 

Median= 𝑛 +21 𝑡ℎ  item 

              =   th item 

        = 5th item 

        = 14 

Median= 14 

Mode=13(largest frequency) 

Range = Largest item -smallest item  

            = 21-13 

            = 8 

5. In a class of 5 students, average weight of the 4 lightest students is 

40 kgs, Average weight of the 4 heaviest students is 45 kgs. What is the difference between the maximum and minimum possible average weight overall? 

Ans. Let a, b, c, d and e the weight of the 5 students such as that a b   c d e.

Since the total weight of the 4 lightest student is 40 kgs*4=160kgs, we have a+ b +c +d 

Similarly, since the total weight of the 4 heaviest student is 45kgs*4=180kgs, we have b+ c+ d+ e 

The maximum average weight of all 5 students occurs when ‘a’ is as larger as possible. However, since the average of the 4 lightest students is 40 kgs, ‘a’ is at most 40, so the maximum average weight is: 

= (a+ b+ c+ d+ e)/5 kgs 

= (40+180)/5 kgs 

= 220/5 kgs 

=44 kgs 

The minimum average weight of all 5 students occurs when ‘e’ is as larger as possible. Since the average of the 4 heaviest students is 45 kgs, ‘e’ is at most 45 kgs, so minimum average weight is: 

= (a+ b+ c+ d+ e)/5 kg 

= (160+45)/5 kg 

=205/5 kgs 

=41 kgs 

Therefore, the difference between the maximum and minimum possible average weight= 44 kg – 41 kg

                            = 3 kg 

6. Find IQR of 3, 5, 1, 8, 6, 2, 6, 4, 9, 4 

Ans: Arrange In ascending order 1,2,3,4,4,5,6,6,8,9 

N=10 

Q1= sizes of (n+1)/4 th term  

    =size of (10+1)/4 th term 

    = size of 2.75th term 

   = 2ndterm + .75(3rdterm-2nd term) 

    =2+.75(3-2) 

  =2.75 

Q3=sizes of 3(N+1)/4 th term  

     = sizes of 3(10+1)/4 th term  

     =sizes of 3*2.75th term  

     =sizes of 8.25th term 

     = 8th term + .25(9th term- 8th term) 

     =6+.25(8-6) 

     =6.5 

INQ=Q1-Q3 

        = 6.5-2.75 

     = 3.75 

7.Find INQ of 21,29,31,45,62,75,25,78,99 

Ans: Arranging in ascending order 21, 25, 29,31, 45, 62, 75, 78, 99 

N=9 

Q1= sizes of (N+1)/4 th term 

     = sizes of (9+1)/4 th term 

     = sizes of 2.5th term 

     = 2nd term + .5(3rd term-2nd term) 

     = 25 +.5(29-25) 

     = 25 + 2 

     =27 

Q3= sizes of 3(N+1)/4 th term 

     = sizes of 3(9+1)/4 th term 

     = sizes of 7.5th term 

    = 7th term + .5 (8th term- 7th term) 

    = 75 + .5 (78-75) 

    = 76.5 

1. A shopkeeper has 50 cold drink bottles. Some of bottles are 1-litre and some are 2-litre bottles. The average cold drink of the bottles is 1200 ml. Find the number of 2-liter bottles. 

Ans: A shopkeeper cold drink bottles=50 

Let as number of 1-liter bottles=X 

And the number of 2-liter bottles=Y 

Therefore X+Y=50……………. (i) 

The average cold drink of the bottles is 1200 ml or 1.2-liter 

Therefore X*1-liter + y*2-liter/50=1.2-liter 

X+2Y=1.2*50 

X+2Y=60………… (ii) 

After solving the two equations, I get X=40, Y=10 

 The number of 2-litre bottles= 10 

Or

Let’s taking no of 1liter bottles=Yand no of 2liter bottles=50-Y=Z

Now, Y×1+Z×2=1.2

Now substitute the value of z in terms of Y

Y×1+2(50-Y) =1.2×50

100-Y=60

Y=40& Z=10

2.A sequence consists of 7 terms arranged in descending order. The mean value of sequence is 70. If 30 is added to each term, and then each term is divided by 2 to get the new mean as ‘K’. Find the difference between K and the original mean. 

Ans: The mean value of sequence =70 

N=7 

If 30 is added to each term, then new mean =70+30 

  =100 

Again if 2 divided by each term, then new mean(K)=100/2 

                                                                                            =50 

The difference between original mean and K=70-50 

        =20 

3. A sequence consists of 9 terms. The standard deviation of the sequence is 50. If 10 is added to each term, and then each term is multiplied by -2. Find the new S.D. 

Ans: The standard deviation of a sequence =50 

If 10 is added to each term, then the standard deviation is same because 10 is added each term 

Again -2 multiplied to each term, then new S.D =50*|2| 

  =100 

The new standard deviation=100 

4. A student has got the following grades on his tests: 87,95,76, and 88. He wants an 85 or better overall. What is minimum grade he must get on the last test in order to achieve that average? 

Ans: A student has got the following grades on his tests= 87,95,76, and 88 

Total number of tests=5 

He wants to achieve the average= 85 or batter 

Minimum total grade he wants=85*5 

  =425 

He got grades in his tests=87,95,76,88 

Total grade he got=87+95+76+88 

     =346 

The minimum grade he must get on the last test in order to achieve that average=425-346 

             =79 

5.In a sequence of 25 terms, can 20 terms be below the average? Can 20 terms be between median and average? 

Ans. Yes, 20 terms can be below the average. No, 20 terms can’t be between median and average.