Guidelines to solve problems chapter 3 Nutan s. Mishra

Exercise 3.52 Variable x = # hours spent partying by a student. Given data for 10 such students. Range = max value of x – min value of x = 14 – 0 = 14 Σx = 64, (Σx) 2 = 4096, Σx 2 = 580 Variance = σ 2 = = Standard deviation =√17.04 = hours

Exercise 3.90(a) Variable x = weight in pounds lost by a member of a health club Ordered dataset is (15+1)/2 the value is Q2 i.e. 8 th value is Q2 i.e. Q2 = 10 pounds Q3 is median of higher seven values Q3 = 14 lbs Q1 is median of lower seven values Q1 = 8 lbs. Q2Q3Q1

3.22, 3.94(a) Variable x = # students suspended /week Size of data set = 12 Data set arranged in increasing order as follows 3, 5, 6, 6, 7, 9, 9, 10, 11, 12, 14, 15 Mean =  x/12 = 107/12 = , Median = mean of 6 th and 7 th values = (9+9)/2 = 9 Mode = 6 and 9 are two modes of this dataset. i.e. this is a bimodal dataset. Q1 = median of lower six values = (6+6)/2 =6 Q3 = median of higher six values =(11+12)/2 = 11.5 IQR = Q3 – Q1 = 11.5 – 6 = 5.5

Exercise 3.23 Variable x = # news papers published in a western state in the year 2000 Size of dataset = 13 Dataset arranged in increasing order is as follows 6, 6, 8, 7, 9, 11, 12, 16, 18, 19, 24, 29, 92 Mean =  x/13 = 257/13 = 19.77, Median is 7 th value in the ordered dataset = 12 In this data set 92 is an outlier value since it is too large compared to rest of the values. Thus after drooping the outlier, new mean =165/12 =13.75 and new median = mean of 6 th and 7 th values =(11+12)/2 =11.5 Since the mean is much bigger than the new mean where as there is not much difference between median and new median.We prefer median as a measure of central tendency in this case.

Exercise X = distance driven during the past year by a sample of drivers in a city (thousands miles) Size of dataset = n= 15 drivers Ordered data set : Mean = Variance = Standard deviation= Range = = Median = Q1 = 10000, Q3 = 26000, IQR = 16000