Minimum 40 maximum 95 range maximum value minimum 55

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Minimum = 40 Maximum = 95 Range = Maximum value – Minimum Value = 95-40 = 55 K stands for the number of classes = 6 Length of the class = 10 Score of an exam frequenc y Relative frequenc y Cumulati ve frequenc y Relative cumulative frequency 40-49 2 0.06 2 0.06 50-59 4 0.133 6 0.2 60-69 7 0.233 13 0.433 70-79 9 0.3 22 0.733 80-89 5 0.166 27 0.9 90-99 3 0.1 30 1 - Total =30 - - -
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3. Calculate the arithmetic mean of the marks from the following table: marks No. of students (frequency) Class midpoint 0-10 12 5 10-20 18 15 20-30 27 25 30-40 20 35 40-50 17 45 50-60 16 55 We compute the mean by using this formula ¯ x = ( f x ) ∑ f where x = midpoint (F*x)= 12*5=60 18*15=270 27*25=675 20*35=700 17*45=765 16*55=880 ( f x ) = 3350 ∑f = 110 ¯ x = ( f x ) ∑ f = 3350 110 = 30.45 4. Construct a stem-and Leaf of the following data; 73, 81, 75, 71, 75, 65, 78, 40, 66, 80, 90, 82, 75, 55, 66. 44, 33, 12, 27, 42. 1 2 2 7 3 3 4 0 2 4 5 5 6 5 6 6 7 1 3 5 5 5 8 8 0 1 2 9 0
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5. An analysis of monthly wages paid to the worker in two firm A and B belong to the Same industry gives the following results: In which firm, A or B is there greater variability in individual wages? Firm A Firm B Number of workers 500 600 Average monthly wage SR. 186.00 SR. 175.00 Variance of distribution of wages 81 100 We compare both of the coefficient variance of the two firms CV = s x *100 SA= 81 = 9 XA = 186 CV for frim A = 9 186 100 = 4.838 SB = 100 = 10 XB = 175 CV for frim B = 10 175 100 = 5.714 5.714 > 4.838 so ,FirmBhas agreater variability individualwages 6. Construct histogram of the following frequency table
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0 2 4 6 8 10 12 14 16 18 Hours of sleep Ferquencey 5.5 10.5
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