Question

Consider the following stem and leaf display for a set of sample data.

Stem Leaves (unit = 0.01)

0 |9

1 |5 6

2 |1 a 8 9

3 |2 3 5 7 8

4 |4 5 b 9

5 |2 3 9   
  (a)

(i) Find the 1st Quartile (?1) and 3rd Quartile (?3) in terms of a and b. (2 marks)

(ii) Suppose the inter-quartile range is 0.22 and the distances between the two quartiles and the median are the same (i.e. Q3- Median = Median-Q 1 ). Find the values of a and b. (2 marks)

(b) Suppose there is at least 75% data within the interval X to Y. By using the Chebyshev’s theorem, find the interval X to Y around the sample mean. (Correct your final answers to 3 decimal places) (4 marks)

(c) Suppose a frequency table for the above data set is constructed with 5 classes. The smallest data value from the sample is used as the lower class limit of the first class. Calculate the mean and variance of the grouped data using the frequency table. (Correct your final answers to 3 decimal places) (4 marks)

Answer 1

Answer:

The stem-and-leaf plot for a sample of 19 observations is provided where the leaf unit is 0.01. The minimum value in the data-set is 0.09 and the maximum value is 0.59. (2.1) The first quartile (Q) is calculated as shown below: 2=(19:42 ) " observation observation = 5observation = 0.2a The third quartile (Q) is calculated as shown below: O: =(349+1) " observation = (15)" observation = 0.46 Therefore, the 1* quartile and 3rd quartiles are 0.2a and 0.46), respectively.

(a. 11) According to the question, the interquartile range is 0.22 and Q - Median = Median - Q. The median is calculated as follows: Median observation = ( n+) observation - (1974) "observation = 10 observation = 0.35 The interquartile range is calculated as follows: Interquartile range = Q - Q. 0.22 =0.46 -0.2a Q.-Median = Median - Q, is resolved as follows: Q: - Median = Median - Q Q-0.35 =0.35- 0,+Q=0.35 +0.35 0.4b +0.2a=0.70

By adding the equations, 0.22 = 0.45-0.2a and 0.4b+0.2a =0.70, the value of b is determined as follows: 0.45 +0.2a +0.45-0.2a=0.70 +0.22 2 x 0.4b = 0.92 0.4b =0.46 6=6 Similarly, the value of a is determined as follows: 0.46 +0.2a -0.45 +0.2a=0.70 -0.22 2 x 0.2a=0.48 0.2a=0.24 a=4 Hence, the value of a and b are 4 and 6, respectively.

(6) According to the Chebyshev's inequality, 75% of the data lies between the two-standard deviation of the mean. That is, the cut off values within which 75% of the data lies is ( 20) where is the sample mean and o is the standard deviation The sample mean is calculated as follows: 1 0.09 +--+0.59 19 =0.35 The standard deviation is calculated as follows: o= x2(3 - 2) = [(0.09 -0.35)*+-+(0.59-0.35)] = 0.137 Therefore, the interval X to Y is determined as follows: (X,Y)=(+20) =(0.352x0.14) = (0.076.0.624)

The class width is calculated as follows: Total number of class = Maximum value-Minimum value Class width 0.59-0.09 Class width Class width = 0.10 The lower limit of the first class is 0.09. 0.010 is added with 0.09 to get the lower limit of the next class and the process is continued till the lower limit of fifth class is obtained. The lower limits of all five classes are 0.09. 0.19.0.29. 0.39, and 0.49. The upper limit of first class can be determined by lower limit of the second class. Then, the class width is added to get the upper limits of other classes. The upper limits are 0.19, 0.29, 0.39.0.49, and 0.59. Frequency of each class can be determined by counting the number of observations whose length lies within a particular class limit. The frequency distribution is obtained as follows: Frequency Length 0.09-0.19 0.19-0.29 0.29-0.39 0.39-0.49 0.49-0.59

The necessary calculation for obtaining the mean is done using Excel software. The screenshot is shown below. 1 Lower limit Upper limit Midpoint (Lower limit+Upper limit 2 Frequency Frequency Midpoint 2 0.09 0.19 -(A2B2)2 -D2°C2 3 0.19 10.29 1-(A3+B3)2 13 -D3°C3 4 0.29 0 39 =(A4+B-42 -D4°C4 5 0.39 10.49 -(AS+B5)2 J-DSCS 60.419 0.59 (A6+B6) 2 -D6°C6 Therefore, the mean length of the screw is calculated as follows: (Frequency x Midpoint) Mean: Total frequency 0.42+ +2.16 6.66 19 =0.351

The calculation for obtaining the mean is done using Excel software. The screenshot is shown below. 1 Lower limit Upper limit Midpoint-(Lower limit+Upper limit 2 Frequency FregenerMidpoint Frequency (Midpoint-0.351) 2 2 0.09 0.19 A2+B2)/2 102"C2 D2*(C2-0.350/12) 3 0.19 0.29 (A3+B3Y2 -D3*C3 -D3(C3.0 351/2) 4 0.29 0.39 (AB4Y2 6 DI"CH DI*(C4-0.3512) 5 0.39 0.49 (AS B92 -DSCS -DS(C5-0.3512) 6 0.49 0.59 (A6B6)/2 16°C6 -D6C6-0.35172) Hence, the standard deviation is calculated as follows: Standard deviation = equency (Midpoint - Mean)) (0.134+..+0.143) 0.338 19 = 0.133 Hence, the mean and standard deviation of the grouped data are 0.351 and 0.133, respectively.