Question

1. ABC toy factory has 10,000 workers. The salaries of the workers are based on the amount of toys they can make in each month. A random sample of 100 workers and the number of toys they have made in a week is selected and the results are tabulated below.

Toys Pieces Number of workers

101 − 150 6

151 − 200 13

201 − 250 15

251 − 300 24

301 − 350 25

351 − 400 9

401 − 450 8

(a) Calculate the mean, median, mode and standard deviation of the number of pieces of toys made by this sample of 100 workers. [ 8 marks]

(b) Estimate from the frequency distribution table, the proportion of workers having a weekly production between 180 and 220 pieces of toys. [5 marks]

(c) Construct a 98% confidence interval for proportion of all workers who can make between 170 and 280 pieces of toys.

Answer 1

Solution:

Class (1)	Frequency (f) (2)	Mid value (x) (3)	f⋅x (4)=(2)×(3)	f⋅x2=(f⋅x)×(x) (5)=(4)×(3)	cf (6)
101-150	6	125.5	753	94501.5	6
151-200	13	175.5	2281.5	400403.25	19
201-250	15	225.5	3382.5	762753.75	34
251-300	24	275.5	6612	1821606	58
301-350	25	325.5	8137.5	2648756.25	83
351-400	9	375.5	3379.5	1269002.25	92
401-450	8	425.5	3404	1448402	100
---	---	---	---	---	---
--	n=100	--	∑f⋅x=27950	∑f⋅x2=8445425	--

Mean ˉx=

=27950/100

=279.5


To find Median Class
= value of (n/2)th observation

= value of (100/2)th observation

= value of 50th observation

From the column of cumulative frequency cf, we find that the 50th observation lies in the class 251-300.

∴ The median class is 250.5-300.5.

Now,
∴L=lower boundary point of median class =250.5

∴n=Total frequency =100

∴cf=Cumulative frequency of the class preceding the median class =34

∴f=Frequency of the median class =24

∴c=class length of median class =50

Median M=L+ *c

=250.5+{[50-34]/24}*50


=250.5+33.3333

=283.8333


To find Mode Class
Here, maximum frequency is 25.

∴ The mode class is 300.5-350.5.

∴L=lower boundary point of mode class =300.5

∴f1= frequency of the mode class =25

∴f0= frequency of the preceding class =24

∴f2= frequency of the succedding class =9

∴c= class length of mode class =50

Z=L+ *c

=300.5+([25-24]/2⋅25-24-9)⋅50

=300.5+(117)⋅50

=300.5+2.9412

=303.4412


Sample Standard deviation S=


=


=79.9874

(b)the proportion of workers having a weekly production between 180 and 220 pieces of toys is (13+15)/100 = 0.28

(c) the proportion of workers who can make between 170 and 280 pieces of toys () =  (13+15+24)/100 = 52/100 = 0.52

n = 100

z at 98% is 2.33

98% confidecen interval = = =

= (0.4038 ,0.6362)