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Mathematics

The table, given below, shows the frequency distribution of the weekly wages of the employees of a company :

Weekly Wages (in ₹)Number of employees
800 - 89922
900 - 99927
1000 - 109923
1100 - 119918
1200 - 129915

Find :

(i) the lower limit of the fourth class.

(ii) the upper limit of the fifth class.

(iii) the class boundaries of the second class.

(iv) the class mark of the first class.

(v) the class size of the third class.

(vi) cumulative frequency of the fourth class.

Statistics

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Answer

(i) The fourth class is 1100 - 1199.

Hence, lower limit of the fourth class is 1100.

(ii) The fifth class is 1200 - 1299.

Hence, the upper limit of the fifth class is 1299.

(iii) The second class is 900 - 999.

The adjustment factor =

=Upper limit of current classLower limit of previous class2=9008992=12=0.5= \dfrac{\text{Upper limit of current class} - \text{Lower limit of previous class}}{2}\\[1em] = \dfrac{900 - 899}{2}\\[1em] = \dfrac{1}{2}\\[1em] = 0.5\\[1em]

Lower boundary = 900 - 0.5 = 899.5

Upper boundary = 900 + 0.5 = 999.5

Hence, the class boundaries of the second class are 899.5-999.5.

(iv) The first class is 800 - 899.

Class mark = lower limit + upper limit2\dfrac{\text{lower limit + upper limit}}{2}

= 800+8992\dfrac{800 + 899}{2}

= 16992\dfrac{1699}{2}

= 849.5

Hence, the class mark of the first class = 849.5.

(v) The third class is 1000 - 1099.

The adjustment factor =

=Upper limit of current classLower limit of previous class2=10009992=12=0.5= \dfrac{\text{Upper limit of current class} - \text{Lower limit of previous class}}{2}\\[1em] = \dfrac{1000 - 999}{2}\\[1em] = \dfrac{1}{2}\\[1em] = 0.5\\[1em]

The actual lower limit = 1000 - 0.5 = 999.5

The actual upper limit = 1099 + 0.5 = 1099.5

Class size = Actual upper limit - Actual lower limit

= 1099.5 - 999.5

= 100

Hence,the class size of the third class = 100.

(vi) Cumulative frequency up to the fourth class = 22 + 27 + 23 + 18 = 90

Hence, cumulative frequency of the fourth class = 90.

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