Question

73 Wed Aug 7 0:43 0 webgunet My Blackboard Content - Blackboard Learn Homework 23 Math 141 Sect. 8-3) - Math 141 section 2015 5. -12 points 0/6 Submissions Used My Notes The following table gives the math SAT scores for males and females in the past five years. Find the average (mean), standard deviation, and variance for each group (Round your answers to two decimal places.) Males 497 500 499 501 502 Females 455 450 452 453 448 Males: mean standard deviation variance Females: mean standard deviation variance

Answer 1

We have given : Male and Female variable and each have five observation

for Male : X

Mean =sum(Xi) / n

=(497+500+499+501+502) / 5

=2499 / 5

= 499.8

Standard deviation = SQRT (VAR(X))     for sample

VAR (X) = (1/(n-1)) * sum (Xi- Mean)^2

=(1/4) *[(497- 499.8)^2 + (500 - 499.8)^2 + (499 - 499.8)^2 + (501 - 499.8)^2 + (502- 499.8)^2 ]

= (1/4) * [(-2.8)^2 + (0.2)^2 + (-0.8)^2 + (1.2)^2 + (2.2)^2]

=(1/4) * [7.84 + 0.04 + 0.64 + 1.44 + 4.84 ]

=(1/4 ) * 14.8

= 3.7

standard deviation = SQRT (3.7) = 1.92

variance = 3.7

#### for Female = Y

Mean =sum(Yi) / n

=(455 + 450 + 452 + 453 + 448) / 5

=2258 / 5

= 451.6

Standard deviation = SQRT (VAR(Y))     for sample

VAR (Y) = (1/(n-1)) * sum (Yi- Mean)^2

=(1/4) *[(455- 451.6)^2 + (450 - 451.6)^2 + (452 - 451.6)^2 + (453 - 451.6)^2 + (448- 451.6)^2 ]

= (1/4) * [(3.4)^2 + (-1.6)^2 + (0.4)^2 + (1.4)^2 + (-3.6)^2]

=(1/4) * [11.56 + 2.56 + 0.16 + 1.96 + 12.96 ]

=(1/4 ) * 29.2

= 7.3

standard deviation = SQRT (7.3) = 2.70

variance = 7.3

Here Female variance is more than Male variance hence group for Female is more Varying than group Male.