{"id":3473,"date":"2023-10-05T10:36:33","date_gmt":"2023-10-05T10:36:33","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3473"},"modified":"2023-10-05T10:42:53","modified_gmt":"2023-10-05T10:42:53","slug":"week-9-ss2-third-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-9-ss2-third-term-further-mathematics-notes\/","title":{"rendered":"Week 9 – SS2 Third Term Further Mathematics Notes"},"content":{"rendered":"

WEEK NINE
\n<\/strong>TOPIC : APPLICATION OF INTEGRATION II : SOLID REVOLUTION AND TRAPEZOIDAL RULE
\n<\/em><\/strong>A solid whichb has a central axis of symmetry is a solid of revolution.<\/em><\/strong>Forexample, a cone, a cylinder , a vase etc.
\n<\/strong>
\n\u00a0\"\"\/
\n\t\t\t<\/strong>y
\n<\/strong>
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0Consider the area under a portion AB of the curve y = f(x) revolved about the x axis through four right angle or 3600, each point of the curve describes a circle centered on the x axis. A solid revolution can be thought of as created in this way with the circular plane ends cutting the x \u2013 axis at x = a and x = b.<\/p>\n

\u00a0Let v be the volume of the solid for x = a up to an arbitrary value of x between a and b. Given abincreament dx in x , and y takes an increamentdy and v increases by dv.
\n\"\"\/<\/p>\n

\u00a0\"\"\/<\/p>\n

\u00a0
\n\u00a0
\n\u00a0
\n\u00a0The figure shows a section through the x axis, from this it is seen that the slice dv of bthickness dx is enclosed between two cylinders of outer radius y +dy and inner radius y .
\nThen ,\u03c0y2dx< dv < \u03c0 (y + dy)2dx;
\nWith appropriate modification, if the curve is falling at this point.
\n\"\"\/\u03c0 y2 < dv\/dx < \u03c0 (y + dy)2
\n\"\"\/\"\"\/\"\"\/if dx 0, dy 0 as dv\/dx dv\/dx
\n: . dv\/dx = \u03c0y2 or V = <\/p>\n

\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0Where y = f(x) and v = volume of solid revolution of the curve where y = f(x) is rotation completely and x \u2013 axis between limits x = a and x =b.<\/p>\n

\u00a0Examples;The portion ofthe curve y = x2<\/sup> between x = 0 and x = 2 is rotated complrtely around the x axis, find the volume of the solid generated?
\n\"\"\/
\n\t\tV =
\n=
\nV =
\nV = =
\nPut x = 2 and x = 0 then substitute into the expression above
\nV = <\/p>\n

\u00a0THE TRAPEZOIDAL RULE
\n<\/strong>There are many definite integrals which can’t be evaluated and thus required advance techniques e.g
\netc
\n\"\"\/We can find an approximate value for such integralsbyb finding the area approximately. There are many methods methods of doing this and such methods include the Trapezium rule.<\/em><\/strong>
\n\t\t Y<\/p>\n

\u00a0
\n\u00a0
\n\u00a0
\n\u00a0 Y = f(x)<\/p>\n

\u00a0 y1<\/sub> y2<\/sub>y3<\/sub> yn-1<\/sub>yn<\/sub>
\n\t\tx1<\/sub> x2<\/sub> x3<\/sub> xn-1<\/sub>xn<\/sub><\/p>\n

\u00a0
\n\u00a0
\n\u00a0 = \u00bd (y1<\/sub> +y2<\/sub>)h + \u00bd (y2<\/sub> +y3<\/sub>)h + \u00bd (yn-1<\/sub> +yn<\/sub>)h
\n\u00bd h(y1+2y2+2y3+\u2026\u2026\u2026.+2yn-1 + yn)
\n\u00bd (width of each trap. ) \u00d7 (first ordinate + last ordinate )+ 2( sum of all other ord.)<\/p>\n

\u00a0F(x)dx = \u00bd h{y1 + yn} +2{y2 +y3 + \u2026Yn}.<\/p>\n

\u00a0Example
\n<\/strong>Find the approximate value of at interval 0.5<\/p>\n

\u00a0<\/p>\n

\n\n\n\n\n
X<\/td>\n1<\/td>\n1.5<\/td>\n2<\/td>\n2.5<\/td>\n3.0<\/td>\n<\/tr>\n
Y = 1\/x <\/td>\n1<\/td>\n0.67<\/td>\n0.5<\/td>\n0.4<\/td>\n0.33<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

\"\"\/
\n\t\tApplying the rule;
\n{ \u00bd . \u00bd { (1 +0.33) + 2 ( 0.67 + 0.5 + 0.4)}
\n\u00bc {(1.33) +2(1.57)}
\n\u00bc (4.47) = 1.12. <\/p>\n

\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0 1 0.67 0.5 0.4 0.33<\/p>\n

\u00a0Ex (2). Make a table of value of y for which y = for which x =2 t0 x = 3 at interval of 0.2.<\/p>\n

\n\n\n\n\n\n\n
X<\/td>\n2<\/td>\n2.2<\/td>\n2.4<\/td>\n2.6<\/td>\n2.8<\/td>\n3<\/td>\n<\/tr>\n
X2-1<\/td>\n3<\/td>\n3.84<\/td>\n4.76<\/td>\n5.76<\/td>\n6.84<\/td>\n8<\/td>\n<\/tr>\n
<\/td>\n1.732<\/td>\n1.956<\/td>\n2.182<\/td>\n2.4<\/td>\n2.615<\/td>\n2.828<\/td>\n<\/tr>\n
<\/td>\n0.5754<\/td>\n0.5703<\/td>\n0.4583<\/td>\n0.4167<\/td>\n0.3824<\/td>\n0.3536<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

\u00a0Using the rule.
\n= \u00bd *0.2 (0.5774 +0.3536 + 3.5354)
\n0.44664 *2
\n0.89 correct to 2 dp<\/p>\n

\u00a0APPLICATION OF INTEGRATION TO KINEMATICS
\n<\/strong>If the ve;locityis given as a function of time, the displacement is the integral of the velocity function with respect to the time .
\nds\/dt = f(t)
\nthen S =
\n= f(t) + C
\nSimilarly, if the acceleration is a function of time, the velocity is the integral of the acceleration function.
\nEx. A particle is projected in a straight line from O until a speed of 6m\/s is attained. At time t secs.Later,its acceleration is (1 + 2t) m\/s2<\/sup> for the value of t = 4. Calculate for the particle (i) its velocity (ii) its distance from O<\/p>\n

\u00a0dv\/dt = 1+ 2t
\nv = = t + t2<\/sup> + C
\nwhen t = 0
\nv = 6m\/sand c = 6.
\nV = (t2<\/sup> + t + 6) m\/s
\nWhen t = 4, v = 16 + 4 + 6 = 26m\/s.
\n(ii) distance (s) = ds\/dt = t2 + t + 6
\nS =
\nS = {t3\/3 + 16\/2 6t}4<\/sup>
\n\t\t = 160\/3
\nm. <\/p>\n

\u00a0Evaluation
\n<\/strong>1. Find the area enclosed byb y = x2 \u2013 x -2 and the x axis
\n2. find the area under the curve y = x2\/3 between x = 2 and x = k is 8 times the area under the same curve between x = 1 and x =2, hence find the value of k.<\/p>\n

\u00a0GENERAL EVALUATION
\n<\/strong><\/p>\n

    \n
  1. A particle moves in astraight line from O until the initial velocity was 2m\/s. its acceleration is given by (2t -3)m\/s2. Calc. (i) its velocity after 3 secs. (ii) the distance from O when it is momentarily at rest.\n<\/li>\n
  2. \n
    Find the volume of solid revolution when a is the region bounded by the cuerve y = 2x. and the ordinate at x = 2,and x = 4 and the x axis is revolved by 2\u03c0.\n<\/div>\n

    \u00a0Reading Assignment<\/strong> :F\/Matrhs Project, pg 47 \u2013 63<\/p>\n

    \u00a0WEEKEND ASSIGNMENT
    \n<\/strong>1. Integrate 2\u221ax A. B.4x3\/2 <\/sup>+ C C. + C D.
    \n2. Integrate A. + C B. x3\/2 <\/sup>+ C C. x2<\/sup> + C D. + C
    \n3. The gradient of a curve is 6x + 2 and it passes through the point (1,3), find its equation A. 3x2<\/sup> – 2x + 2 B. 3x2<\/sup> -2x -2 + 2x + 2 C. 3x2<\/sup> \u2013 2x + 2 D. 3x2<\/sup> \u2013 2x -2
    \n4. Evaluate A. \u2013 + x2 \u2013 2x + C B. x3\/3 \u2013 x3\/2 + x2 + 2x +C C. x3\/3 + x2\/2 +x3-2x + C D. x4\/4+x3\/3-x2+2x+C
    \n5. Eval. A. 9 +C B. 8\/3 + C C. 24 D. 18 + C
    \nTHEORY
    \n<\/strong><\/li>\n

  3. Evaluate\n<\/li>\n
  4. \n
    Using trapezoidal rule, with ordinate x = -3,-2,-1,0,1,2,3 and 4. Calc correct to 3 dp an approximate value of\n<\/div>\n

    \u00a0
    \n\u00a0WEEK TEN
    \n<\/strong>TOPIC: REGRESSION LINE AND CORRELATION COEFFICIENT
    \n<\/strong>SCATTER DIAGRAM
    \n<\/strong>Definition:<\/strong> a scatter diagram is a graphic display of bivariate data. A bivariate data involves two variables
    \nTYPES OF SCATTER DIAGRAM:
    \n<\/strong>Linear positive correlation.
    \n<\/strong>A positive correlation between two variables x any y means that in general, increase in x is accompanied by increase in y. The regression line has a positive slope.
    \n\"\"\/\"\"\/<\/p>\n

    \u00a0
    \n\u00a0
    \n\u00a0X<\/p>\n

    \u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0Linear negative correlation
    \n<\/strong>A negative correlation between x and y means that an increase in x is accompanied by a decrease in y, negative correlation has a negative slope.
    \n\"\"\/\"\"\/<\/p>\n

    \u00a0
    \n\u00a0
    \n\u00a0x<\/p>\n

    \u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0\"\"\/Zero Correlation: <\/strong>
    \n\t\t\t\t\"\"\/There is no apparent association between x and y.
    \ny<\/p>\n

    \u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0Non Linear Correlation:<\/strong>
    \n\t\t\t\t\"\"\/Most of the points lie on or near a curve which is parabolic in shape. The parabolic curve is called a regression curve.<\/p>\n

    \u00a0
    \n\u00a0
    \n\u00a0x<\/p>\n

    \u00a0
    \n\u00a0
    \n\u00a0
    \n\u00a0REGRESSION LINE OR LINE OF BEST FIT OR THE LEAST SQUARES LINE <\/strong>
    \n\t\t\t\tThere are two variables where one is dependent and the other is independent variable. The regression line can be fit using scatter diagram method and the least squares method.<\/p>\n

    \u00a0LEAST SQUARES METHOD<\/em><\/strong>: If x is independent variable and y dependent variable, that is y on x. then :The equation of the regression line is written as y = ax + b
    \nWhere a is the slope and b is the y \u2013 intercept. Given two sets of variables x and y it can be deduced that
    \na = n \u2211 xy \u2013 \u2211 x \u2211 y
    \n<\/strong>\"\"\/ \u2211 x2<\/sup> \u2013 ( \u2211 x)2<\/sup>
    \n\t\t\t\t\t<\/strong>\"\"\/\"\"\/b = y a \u2013 ax
    \n\"\"\/\"\"\/Where x = \u2211 x
    \nn<\/p>\n

    \u00a0\"\"\/y = \u2211 y
    \n\"\"\/n
    \nExample:<\/strong> use the least square method to fit a regression line of y on x for the following data <\/p>\n

    \n\n\n\n\n
    X<\/td>\n3<\/td>\n5<\/td>\n6<\/td>\n9<\/td>\n11<\/td>\n14<\/td>\n15<\/td>\n18<\/td>\n<\/tr>\n
    Y<\/td>\n2<\/td>\n3<\/td>\n5<\/td>\n7<\/td>\n10<\/td>\n12<\/td>\n13<\/td>\n17<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

    Find value of y when x = 8<\/p>\n

    \u00a0SOLUTION:
    \n<\/strong><\/p>\n

    \n\n\n\n\n\n\n\n\n\n\n\n\n
    X<\/td>\ny <\/td>\nXy<\/td>\nx2<\/sup><\/td>\n<\/tr>\n
    3<\/td>\n2<\/td>\n6<\/td>\n9<\/td>\n<\/tr>\n
    5<\/td>\n3<\/td>\n15<\/td>\n25<\/td>\n<\/tr>\n
    6<\/td>\n5<\/td>\n30<\/td>\n36<\/td>\n<\/tr>\n
    9<\/td>\n7<\/td>\n63<\/td>\n81<\/td>\n<\/tr>\n
    11<\/td>\n10<\/td>\n110<\/td>\n121<\/td>\n<\/tr>\n
    14<\/td>\n12<\/td>\n168<\/td>\n196<\/td>\n<\/tr>\n
    15<\/td>\n13<\/td>\n195<\/td>\n225<\/td>\n<\/tr>\n
    18<\/td>\n17<\/td>\n306<\/td>\n324<\/td>\n<\/tr>\n
    \u2211 x = 81<\/td>\n\u2211 y = 69<\/td>\n\u2211 xy = 893<\/td>\n\u2211 x2<\/sup>= 1017<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

    \u00a0\"\"\/ a = n \u2211 xy \u2013 \u2211x \u2211 y= 8 (893) – 81x 69
    \n\"\"\/n\u2211(x2 <\/sup>) \u2013 ( \u2211x)2<\/sup> 8 (1017) \u2013 (81)2<\/sup>
    \n\t\t\t\t\"\"\/\u00a0\u00a0\u00a0\u00a0 a =\u00a0\u00a0\u00a0\u00a07144 \u2013 5589 = 1555
    \n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a08136 \u2013 6561 1575
    \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0a = 0. 9873
    \n\"\"\/x = \u2211 x = 81 = 10.125
    \n\"\"\/\"\"\/n 8
    \n\"\"\/y = \u2211 y = 69 = 8. 625
    \nn 8
    \n\"\"\/\"\"\/b = y \u2013 ax
    \n b = 8.625 — 0.9873 (10.125)
    \n = 8.625 \u2013 9.996
    \n b = -1.37
    \n y = ax + b
    \n y = 0.9873x \u2013 1.37 (regression line of y on x )
    \n When x = 8
    \n y = 0. 9873 (8) \u2013 1.37
    \n y = 6.5284 ~ 6. 5
    \nEVALUATION
    \n<\/strong>Use the least square method to fit a regression line of y on x for the following data <\/p>\n

    \n\n\n\n\n
    X<\/td>\n1<\/td>\n4<\/td>\n5<\/td>\n7<\/td>\n8<\/td>\n10<\/td>\n12<\/td>\n16<\/td>\n19<\/td>\n20<\/td>\n<\/tr>\n
    Y<\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n7<\/td>\n8<\/td>\n10<\/td>\n15<\/td>\n20<\/td>\n18<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

    Use the line obtained to find the value of y when x = 9<\/p>\n

    \u00a0CORRELATION COEFFICIENT
    \n<\/strong>DEFINITION:
    \n<\/em><\/strong>The correlation coefficient determines the amount or degree of linear relationship between two variables. The correlation coefficient is represented by r
    \nThe characteristics of r are as follows:\n<\/li>\n

  5. \n
    The value of r is the same irrespective of the variable labelled x or y.\n<\/div>\n<\/li>\n
  6. \n
    the value of r satisfies the inequality -1< x < + 1\n<\/div>\n<\/li>\n
  7. \n
    if r is close to +1, the variables are highly positively correlated. If r is close to -1 then, x and y are highly negatively correlated. If r is close to zero, the correlation between x and y is very low. There is no correlation between x and y when r = 0\n<\/div>\n

    There are two methods of obtaining the correlation coefficient.\n<\/li>\n

  8. \n
    Pearson’s coefficient of correlation or product moment correlation coefficient\n<\/div>\n<\/li>\n
  9. \n
    Rank correlation coefficient.\n<\/div>\n

    \u00a0RANK CORRELATION COEFFICIENT<\/strong>: It is also known as Spearman’s rank correlation coefficient and defined as :
    \n\"\"\/rk<\/sub> = 1 \u2013 6 \u2211 D2<\/sup>
    \n\t\t\t\t\t<\/strong>n(n2<\/sup> -1)
    \n<\/strong>As the name implies, the variables (if not ranked) can be ranked in ascending order or descending order. Where there are ties, the average is used as the rank.<\/p>\n

    \u00a0Where D is the difference between the pairs of variables and n is the number of variables. D = Rx \u2013 Ry
    \nExample:
    \nThe table below gives the examination marks of 10 students in mathematics and history.<\/p>\n

    \n\n\n\n\n
    Maths<\/td>\n51<\/td>\n25<\/td>\n33<\/td>\n55<\/td>\n65<\/td>\n38<\/td>\n35<\/td>\n53<\/td>\n61<\/td>\n44<\/td>\n<\/tr>\n
    History<\/td>\n20<\/td>\n65<\/td>\n25<\/td>\n36<\/td>\n51<\/td>\n50<\/td>\n77<\/td>\n31<\/td>\n60<\/td>\n5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

    \u00a0A\u00a0\u00a0\u00a0\u00a0Calculate the rank correlation coefficient
    \nb) \u00a0\u00a0\u00a0\u00a0Comment briefly on your result<\/p>\n

    \u00a0SOLUTION:<\/p>\n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n
    MATHS (x)<\/td>\nHISTORY (y)<\/td>\nRx<\/td>\nRy<\/td>\nD<\/td>\nD2<\/sup><\/td>\n<\/tr>\n
    51<\/td>\n20<\/td>\n5<\/td>\n9<\/td>\n-4<\/td>\n16<\/td>\n<\/tr>\n
    25<\/td>\n65<\/td>\n10<\/td>\n2<\/td>\n8<\/td>\n64<\/td>\n<\/tr>\n
    33<\/td>\n25<\/td>\n9<\/td>\n8<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
    55<\/td>\n36<\/td>\n3<\/td>\n6<\/td>\n-3<\/td>\n9<\/td>\n<\/tr>\n
    65<\/td>\n51<\/td>\n1<\/td>\n4<\/td>\n-3<\/td>\n9<\/td>\n<\/tr>\n
    38<\/td>\n50<\/td>\n7<\/td>\n5<\/td>\n2<\/td>\n4<\/td>\n<\/tr>\n
    35<\/td>\n77<\/td>\n8<\/td>\n1<\/td>\n7<\/td>\n49<\/td>\n<\/tr>\n
    53<\/td>\n31<\/td>\n4<\/td>\n7<\/td>\n-3<\/td>\n9<\/td>\n<\/tr>\n
    61<\/td>\n60<\/td>\n2<\/td>\n3<\/td>\n-1<\/td>\n1<\/td>\n<\/tr>\n
    44<\/td>\n5<\/td>\n6<\/td>\n10<\/td>\n-4<\/td>\n16<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u2211D2<\/sup> = 178
    \nrk<\/sub> = 1- 6 \u2211D2
    \n<\/sup>\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0n(n2<\/sup> \u2013 1)<\/p>\n

    \u00a0=1 \u2013 6 x 178
    \n\"\"\/ 10 (102<\/sup> \u2013 1)
    \n 1 – 1068\/990
    \n= 1-1.178= -0.078<\/p>\n

    \u00a0There is a very low negative correlation between the marks obtained in mathematics and history.<\/p>\n

    \u00a0EVALUATION:
    \n<\/strong>The table below shows the marks obtained by ten students in both theory (x) and practical (y) examination.<\/p>\n

    \n\n\n\n\n
    X<\/td>\n50<\/td>\n 70<\/td>\n85<\/td>\n35<\/td>\n60<\/td>\n65<\/td>\n75<\/td>\n40<\/td>\n45<\/td>\n80<\/td>\n<\/tr>\n
    Y <\/td>\n45<\/td>\n55<\/td>\n75<\/td>\n40<\/td>\n50<\/td>\n60<\/td>\n70<\/td>\n35<\/td>\n30<\/td>\n65<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

    \u00a0 Calculate the rank correlation coefficient between x and y comment on your result.<\/p>\n

    \u00a0PEARSON’S CORRELATION COEFFICIENT<\/strong>: It is fully called Pearson’s product moment correlation coefficient. It is simple to calculate and it does not recognise any of the variables as independent or dependent. It is obtained using the formula below.<\/p>\n

    \u00a0 r = n \u2211 xy – \u2211x\u2211 y
    \n\t\t\t\t\t\t<\/strong>\"\"\/ \u221a [n\u2211(x2<\/sup> ) \u2013 (\u2211x)2<\/sup> ][n\u2211(y2<\/sup>) \u2013 (\u2211y)2<\/sup>
    \n\t\t\t\t\t<\/strong>Example:\u00a0\u00a0\u00a0\u00a0
    \n<\/strong>\u00a0\u00a0\u00a0\u00a0Calculate the product moment correlation coefficient for the following data <\/p>\n

    \n\n\n\n\n
    X <\/td>\n 2<\/td>\n 4<\/td>\n 7<\/td>\n 9<\/td>\n 11<\/td>\n<\/tr>\n
    Y <\/td>\n 1<\/td>\n 2<\/td>\n 3<\/td>\n 7<\/td>\n 9<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

    Comment on your result.<\/p>\n

    \u00a0SOLUTION:
    \n<\/strong><\/p>\n

    \n\n\n\n\n\n\n\n\n\n
    X <\/td>\n Y <\/td>\n XY<\/td>\n X2<\/sup><\/td>\n Y2<\/sup><\/td>\n<\/tr>\n
    2<\/td>\n 1<\/td>\n 2<\/td>\n 4<\/td>\n 1<\/td>\n<\/tr>\n
    4<\/td>\n 2<\/td>\n 8<\/td>\n 16<\/td>\n 4<\/td>\n<\/tr>\n
    7 <\/td>\n 3<\/td>\n 21<\/td>\n 49<\/td>\n 9<\/td>\n<\/tr>\n
    9<\/td>\n 7<\/td>\n 63<\/td>\n 81<\/td>\n 49<\/td>\n<\/tr>\n
    11<\/td>\n 9<\/td>\n 99<\/td>\n 121 <\/td>\n 81<\/td>\n<\/tr>\n
    \u2211x = 33<\/td>\n\u2211y = 22<\/td>\n \u2211xy = 193<\/td>\n\u2211x2<\/sup> = 271<\/td>\n\u2211y2<\/sup> = 144<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

    \u00a0\"\"\/ r = 5 x 193 \u2013 33 x 22
    \n\u221a[<\/strong>5(271) \u2013 ( 33)2<\/sup>][5(144) \u2013 ( 22)2<\/sup>]
    \n\t\t\t\t\t<\/strong> r = 965 \u2013 726
    \n\t\t\t\t \u221a266 x 236
    \n r = 239
    \n\t\t\t\t 250.55
    \n r = 0.9539. r = 0.95 (approximately to 2 s.f)
    \nComment: The relationship between x and y is highly positive.<\/p>\n

    \u00a0EVALUATION<\/strong>: The following data are the marks obtained by five students in statistics (X) and mathematics(Y). Calculate the product moment correlation coefficient and comment on your result.<\/p>\n

    \n\n\n\n\n
    X<\/td>\n 33<\/td>\n 36<\/td>\n 42<\/td>\n 52<\/td>\n 40<\/td>\n<\/tr>\n
    Y<\/td>\n 42<\/td>\n 46<\/td>\n 38<\/td>\n 62<\/td>\n 52<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

    \u00a0GENERAL EVALUATION\/REVISIONAL QUESTIONS
    \n<\/em><\/strong><\/li>\n

  10. If Cos A = 24\/25 and Sin B= 3\/5, where A is acute and B is obtuse, find without using tables, the values of (a) Sin 2A (b) Cos 2B (c) Sin (A-B)\n<\/li>\n
  11. \n
    Use the addition formula to find the values of the following\n<\/div>\n

    (a)Sin 750<\/sup> (b) cos 750<\/sup> (c) tan 450
    \n<\/sup><\/li>\n

  12. \n
    Calculate the Product moment correlation coefficient and the Spearman’s rank correlation coefficient.\n<\/div>\n
    \n\n\n\n\n
    X<\/td>\n50<\/td>\n45<\/td>\n43<\/td>\n30<\/td>\n30<\/td>\n43<\/td>\n23<\/td>\n43<\/td>\n25<\/td>\n<\/tr>\n
    Y<\/td>\n12<\/td>\n13.5<\/td>\n14<\/td>\n11<\/td>\n12<\/td>\n15<\/td>\n13.5<\/td>\n12<\/td>\n14<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

    \u00a0READING ASSIGNMENT<\/strong>: Read correlation and regression.Page313\u2013320. Further Mathematics project 2.
    \nWEEKEND ASSIGNMENT
    \n<\/strong>Use the table below to answer questions 1 and 2.<\/p>\n

    \n\n\n\n\n
    Height <\/td>\n 160<\/td>\n161<\/td>\n162<\/td>\n163<\/td>\n164<\/td>\n165<\/td>\n<\/tr>\n
    No of students<\/td>\n 4<\/td>\n6<\/td>\n 3<\/td>\n 7<\/td>\n8<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

    \u00a0<\/li>\n

  13. \n
    The mean of the distribution is\n<\/div>\n

    (a) 4875.1 cm ( b) 4001.2 (c) 3571.0cm (d) 162.2 cm (e) 129.2cm
    \n2. \u00a0\u00a0\u00a0\u00a0The median of the distribution is
    \n\u00a0\u00a0\u00a0\u00a0(a) 160 (b) 162 (c) 163 (d) 164 (e) 165
    \n3.\u00a0\u00a0\u00a0\u00a0Calculate the standard deviation of 3,4, 5,6,7,8,9
    \n\u00a0\u00a0\u00a0\u00a0 (a) 2 (b) 2.4 (c) 3.6 (d) 4.0 (e) 4.2
    \n4.\u00a0\u00a0\u00a0\u00a0Calculate the mean deviation of 6 , 8 , 4 , 0 , 4
    \n\u00a0\u00a0\u00a0\u00a0(a) 4 .0 (b) 3.6 (c) 3.0 (d) 2. 8 (e) 2 . 1
    \n5.\u00a0\u00a0\u00a0\u00a0 The table below shows the rank Rx and Ry of marks scored by 10 candidates in an oral and
    \nwritten tests respectively. Calculate the spearman’s rank correlation coefficient of the data.<\/p>\n

    \u00a0
    \n\u00a0<\/p>\n

    \n\n\n\n\n
    Rx<\/sub><\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n6<\/td>\n7<\/td>\n8<\/td>\n9<\/td>\n10<\/td>\n<\/tr>\n
    Ry<\/sub><\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n1<\/td>\n6<\/td>\n5<\/td>\n8<\/td>\n7<\/td>\n10<\/td>\n9\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

    (a)51\/55 b) 6\/55 c)49\/55 d)54\/55 e) 61\/55<\/p>\n

    \u00a0THEORY
    \n<\/strong>1\u00a0\u00a0\u00a0\u00a0The distribution of marks scored in statistics and mathematics by ten students is given in the table below:<\/p>\n

    \u00a0<\/p>\n

    \n\n\n\n\n
    Maths(x <\/td>\n11<\/td>\n20<\/td>\n 23<\/td>\n42<\/td>\n48<\/td>\n50<\/td>\n57<\/td>\n64<\/td>\n80<\/td>\n90<\/td>\n<\/tr>\n
    Stat(y)<\/td>\n26<\/td>\n23<\/td>\n35<\/td>\n46<\/td>\n44<\/td>\n50<\/td>\n50<\/td>\n58<\/td>\n68<\/td>\n70<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

    \u00a0<\/li>\n

  14. \n
    Plot a scatter diagram for the distribution\n<\/div>\n<\/li>\n
  15. \n
    Draw an eye- fitted line of best fit\n<\/div>\n<\/li>\n
  16. \n
    Use your line to estimate the students marks in statistics if his mark in maths is 40\n<\/div>\n

    2.\u00a0\u00a0\u00a0\u00a0The table below gives the marks obtained by members of a class in maths and physics examination<\/p>\n

    \n\n\n\n\n\n
    STUDENTS<\/td>\nA<\/td>\nB<\/td>\nC<\/td>\nD<\/td>\nE<\/td>\nF<\/td>\nG<\/td>\nH<\/td>\nI<\/td>\nJ<\/td>\n<\/tr>\n
    Maths<\/td>\n85<\/td>\n75<\/td>\n59<\/td>\n43<\/td>\n74<\/td>\n69<\/td>\n62<\/td>\n80<\/td>\n54<\/td>\n63<\/td>\n<\/tr>\n
    Physic<\/td>\n92<\/td>\n72<\/td>\n62<\/td>\n48<\/td>\n85<\/td>\n73<\/td>\n46<\/td>\n74<\/td>\n58<\/td>\n50<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

    \u00a0<\/li>\n

  17. \n
    Calculate the product moment correlation coefficient.\n<\/div>\n<\/li>\n
  18. \n
    Comment on your result.\n<\/div>\n

    \u00a0
    \n\t\t\t\t\u00a0<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

    WEEK NINE TOPIC : APPLICATION OF INTEGRATION II : SOLID REVOLUTION AND TRAPEZOIDAL RULE A…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,275],"tags":[],"class_list":["post-3473","post","type-post","status-publish","format-standard","hentry","category-posts","category-third-term-ss2-further-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3473","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=3473"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3473\/revisions"}],"predecessor-version":[{"id":3474,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3473\/revisions\/3474"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3473"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3473"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3473"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}