Solve the problem, please show process and solutions. II. The frequency distribu

Question

Solve the problem, please show process and solutions.

II. The frequency distribution of the ages of the employees of Stark Tech is given below: Agef 50-54 1 45-495 40-44 | 11 35-39 18 30-34 30 25-29 24 20-24 11 1. Find the following: a. mean b. median c. mode of this distribution. (6 pts) 2. Determine: a. Q2 b. D C. P95 of this distribution. (6 pts) 3. Is the median and Q2 of this distribution equal? Are they always equal in all distribution? Why or why not? (3 pts) Determine the: a. quartile deviation 4. (2 pts) b. median deviation from the mode (2 pts) 5. Determine the: a. quartile coefficient of dispersion (2 pts) c. mean coefficient of dispersion from the mode (2 pts) 6. Determine what particular value range contains or includes 99.73 % of the distribution. (2 pts) Note: Show all necessary solutions.

Explanation / Answer

1.

(a) Mean = 3265/ 100 =32.65

(b) Median = L + [n/2 - B]/ G X w

= 30 + [100/2 - 35] / 30 * 5

= 32.5

(c) Mode = L + (fm - fm-1) / [  (fm - fm-1) + (fm - fm+1) ] x w

= 30 + (30 - 24)/ [ ( 30-24) + ( 30 - 18) ] * 5

= 30 + 6/18 * 5

= 31.67

Q.2 Q2 = median = 32.5

(b) D9 = P90

D9 will be 9 *100/10 = 45 which will be belonged to class 40 -44.

D9 = L + (9N/10 - C)/f * w

= 40 + ( 90 - 83)/ 11 * 5

= 43.18

(c) P95 = L + (95 - C)/f * w

P95 = 45 + (95 - 94) / 5 * 5 = 46

Q.3 Median and Q2 are equal here. Yes, they are always equal for all distriubtion and it divides the data into two equal halfs.

Q.4 Quartile deviation = (Q3 - Q1)/2

Q3 = P75 = L + (0.75N - C)/f * w

= 35 + ( 75 - 65)/ 18 * 5

= 37.78

Q1 = P25 = L + (0.75N - C)/f * w

= 25 + ( 25 - 11)/ 24 * 5

= 27.92

so Quartile deviation = (Q3 - Q1)/2 = (37.78 - 27.92)/2 = 4.93

(b) Median deviation from Mode = Mode - median = abs (32.5 - 31.67) = 0.833

Q/5 (a) Quartile coefficient of dispersion = (Q3 -Q1 )/ (Q3 + Q1 ) = (37.78 - 27.92)/(37.78 + 27.92) = 0.15

(b) Mean coeffieicnt of dispersion from the mode = f * (X- Mode) / f = 4828.79/100 = 48.29

Q.6 We have to find that values contain 99.73% of distriubtion.

The value 20 - 50 will contai the 99.74% of the distriubtion.

Age Midpoint (x) f Cum. f x * f (x-xbar)^2 f * (x - xbar)^2 20-24 22 11 11 242 113.4225 1247.648 25-29 27 24 35 648 31.9225 766.14 30-34 32 30 65 960 0.4225 12.675 35-39 37 18 83 666 18.9225 340.605 40-44 42 11 94 462 87.4225 961.6475 45-49 47 5 99 235 205.9225 1029.613 50 -54 52 1 100 52 374.4225 374.4225 sum 100 3265 4732.75

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