In the study of age category in the new school, observation shows that most of the students in the school fall within certain age range according to the ministry of education line up. A total of B students were recorded for the recent senior school examination in 9 groups. The assume mean of the distribution is 19.75.
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--> (I)
--> The assume mean (X) of the distribution(table) is = 19.75
--> The sum of the distribution (FX) = 60Q^2 + 2922
--> The sum of the frequency(F) = 3Q^2 + 163
--> Assume mean (X) = Sum of |FX| / Sum of |F|
--> Assume mean (X) = [ 60Q^2 + 2922 ] / [ 3Q^2 + 163 ]
--> 19.75 = [ 60Q^2 + 2922 ] / [ 3Q^2 + 163 ]
--> Cross multiply to eliminate the fraction on the righr hand side
--> 19.75 x [ 3Q^2 + 163 ] = [ 60Q^2 + 2922 ]
--> 59.25Q^2 + 3219.25 = 60Q^2 + 2922
--> Combine like terms
--> 0.75Q^2 + = 297.25
--> Divide both sides of the equation by 0.75
--> 0.75Q^2 + / 0.75 = 297.25 / 0.75
--> Q^2 + = 297.25 / 0.75
--> Take the square root of both sides of the equation.
--> Sqrt(Q^2) = Sqrt(297.25 / 0.75)
--> Q = Sqrt(297.25 / 0.75)
--> Q = (+,-) Sqrt(396.333333333333)
--> Q = (+,-) 19.9081222955188
--> Q = + 19.9081222955188
--> Q = - 19.9081222955188
--> However, a negative value cannot be used as the solution of
the distribution. Therefore, the only answer/value available for Q
--> is Q = 19.9081222955188
--> is Q = 19.91 Approximate
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--> To the nearest whole number, Q = 20
--> Q = 20 , to the nearest whole number
--> B = 3Q^2 + 163
--> B = 3 x 20^2 + 163
--> B = 3 x 400 + 163
--> B = 1200 + 163
--> B = 1363
--> The total number of students in taking the examination is =
1363
--> Produced by MATCAL Program
--> Group 6 = 3Q^2 + 24
--> Group 6 = 3 x 20^2 + 24
--> Group 6 = 3 x 400 + 24
--> Group 6 = 1200 + 24
--> Group 6 = 1224
--> Group 6 = 1224
--> The total number of students in Group 6 = 1224
--> (II)
--> Variance Table/Data
--> 23 x 4.75
--> 21 x 3.75
--> 28 x 1.75
--> 16 x 2.75
--> 15 x 0.75
--> 1224 x 0.25
--> 12 x 1.25
--> 14 x 2.25
-> B = Total number of all the students for the examination
1] What is the value of Q that represents the number of students that
fall under number 6 and the value of B to the nearest whole number. Also,
how many students are in group 6.
2] What is the variance of the table.
3] Determine the standard deviation of the age group.
4] Draw a pie chart and a bar chart to represent the table.
5] What is the modal age of the group.
6] Create a tally for the age in the table provided.
--> 10 x 5.75
--------------------------------------->
--> 109.25
--> 78.75
--> 49
--> 44
--> 11.25
--> 306
--> 15
--> 31.5
--> 57.5
--------------------------------------->
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--> Standard Deviation Table/Data
--> 23 x 22.5625
--> 21 x 14.0625
--> 28 x 3.0625
--> 16 x 7.5625
--> 15 x 0.5625
--> 1224 x 0.0625
--> 12 x 1.5625
--> 14 x 5.0625
--> 10 x 33.0625
--------------------------------------->
--> 518.9375
--> 295.3125
--> 85.75
--> 121
--> 8.4375
--> 76.5
--> 18.75
--> 70.875
--> 330.625
------------Calculating Variance--------------------------->
--> Produced by MATCAL Program
--> (2)
--> Variance = Sum[F|X - (X)|] / Sum(F)
--> Sum Of the Variance (V) = 702.25
--> Frequency (F) of the distribution = 1363
--> Variance = 702.25 / 1363
--> Variance = 0.515223771093177
--> Variance = 0.515223771093177
--> Variance = 0.52
------------Standard Deviation--------------------------->
--> (3)
--> Standard Deviation = Sum{Sqrt[F|X - (X)|]} / Sum(F)
--> Sum Of S.D. = 1526.1875
--> Frequency (F) of the distribution = 1363
--> Standard Deviation = 1526.1875 / 1363
--> Standard Deviation = 1.11972670579604
--> S.D. = 1.11972670579604
--> S.D. = 1.12
--------------------------------------->
--> Produced by MATCAL Program
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