Statistics

Section 7 .1

Q1 . of the curves in figure 7 .7 looks like a normal curve because a normal curve is a smooth curve with single peak , which is symmetric about the mean and always remain above the x-axis . As of the curves satisfy these conditions , therefore , they are not normal

16 18 and 2

Q3 . Figure 7 .9 has larger standard deviation . Mean of the curve in 7 .9 is 10 and mean of the curve in figure 7 .10 is 4

Q4 . a

b

c

d p

e ) No , a larger mean is by no way always results in a larger standard deviation . There two parameters represent altogether different attributes of the data . While mean is a measure of the central tendency standard deviation is a measure of the spread of the individual data point around the central value or mean . In case , there is a data having high value points its mean will be higher than that of another data having lower value points . At the same time if the data points in the set having higher mean are closely spaced and those in the lower mean data set are widely dispersed , then the higher mean data set will have smaller standard deviation . Thus higher mean is by no way a guarantee for higher standard deviation and vice versa . Just see the values in Q4 a ) and Q4 . d

Q5 . a ) 50 b ) 68 .26 c ) 99 .74

Q6 . a ) 50 b ) 95 .44 c ) 0 .13

Q7 . a ) 50 b ) 50 c ) 68 .26 d ) 95 .44

Q8 . a ) 95 .44 i .e . 954

b ) 68 .26 i .e . 683

c ) 50 i .e . 500

d ) 99 .74 i .e . 997

- 1243 - 36 [1207 1279]

b - 2 1243 - 2 36 [1171 1315]

c - 3 1243 - 3 36 [1135 1351]

Q10 . a ) 15 .87

b ) 84 .13

c ) 850 0 .1587 135

Q11 . a - 3 .15 - 1 .45 [1 .7 4 .6]

b - 2 3 .15 - 2 1 .45 [0 .25 6 .05]

Section 7 .2

Q1 . a ) For scores above mean z is positive . Therefore , Robert , Linda and Juan scored above the mean

b ) For score on the mean , z 0 . Therefore , Joel scored on the mean

c ) For scores below mean , z is negative . Therefore , Jan and Susan scored below mean

z

Robert scored 150 1 .1 20 172

Juan scored 150 1 .7 20 184

Susan scored 150 - 2 20 110

Joel scored 150 0 20 150

Jan scored 150 - 0 .80 20 134

Linda scored 150 1 .60 20 182

Q2 . a ) z (x - therefore , z for

a ) Dr . Taylor z (45 - 51 /2 .5 -2 .4

b ) Mr . Patterson z (72 - 51 /2 .5 8 .4

c ) Dr . Lee z (75 - 51 /2 .5 9 .6

d ) Ms . Simms z (65 - 51 /2 .5 5 .6

e ) Dr . Adams z (33 - 51 /2 .5 7 .2

f ) Dr . Riley z (55 - 51 /2 .5 1 .6

Q3 . z (x - therefore

a ) x 53 means z (53 - 73 /5 -4

x 93 means z (93 - 73 /5 4

Therefore , the desired z interval is -4 z 4

Similarly , b ) z 1

d ) x 73 1 .75 5 i .e . x 81 .75

e ) x 62 .5

f ) 64 x 81 .25

Q4 . z (x - therefore

a ) z 1 .91

c ) 1 .1 z 17 .87

e ) x 32 .7

f ) 18 .64 x 1 .77

b ) z 1 .61

c ) -1 .45 x 1 .45

d ) 3706 x 5907

e ) x 6000

g ) x 2800 corresponds to z - 2 .58 . This highly negative z value implies unusually low number of deers in the said National park

Similarly x 6300 implies z 3 .06 . This high value of z implies unusually large deer population

Q6 . z (x - therefore

a ) z 0 .86

b ) z - 0 .86

c ) - 2 .3 z - 1 .71

d ) x 11332

f ) 7937 x 1

b ) z - 2

c ) - 2 .67 z 2 .33

d ) x 5 .18

f ) 4 .12 x or 3 .67 , which is highly value and implies unusually high RBC count

Q8 . a ) For x1 1250 z (x1 - ?1 (1250 - 1272 /35 - 0 .63

For x2 1250 z (x2 - ?2 (1234 - 1122 /40 2 .8

These are shown in the diagram below

b ) The value x2 is more unusual as it is lying in at the extreme and therefore , its probability is very low

Q9 . Q10 Q11 . Q12

Q13 . Q14 Q15 . Q16 Q17 . Q18 Q19 . Q20

Q21 . Q22 Q23 . Q24

Q25 . Q26 Q27 . Q28

Q29 . Q30 . Q31 Q32

P (z ? 0 0 .50

(z 0 0 .50

(z ? -0 .13 0 .4483

(z -2 .15 0 .0158

Q33 . Q34 . Q35 Q36

p 2

4

6

br F

L

N

V

\

^

f

l

n

70 0 .8849

(z 1 .35 0 .0869 br

(z 2 .17 0 .0150

Q37 . Q38 . Q39 Q40

P (z -1 .2 0 .8849

(z -1 .5 0 .9332

(-1 .2 z ? 2 .64 0 .8808 br

(-2 .2 z 1 .04 0 .8369

Q41 . Q42 Q43 . Q44

P (-2 .18 z ?-0 .42 )

(-1 .78 z -1 .23 )

(0 z ? 1 .62 br

(0 z 0 .54

0 .323 0 .0718 0 .4477 0 .2054

Q45 . Q46 . Q47 Q48

P (-0 .82 z ? 0 )

(-2 .37 0 )

(-0 .45 z 2 .73 )

(-0 .73 z 3 .12

0 .2939 0 .4911 0 .6704 0 .7664

Section 7 .3

Q1 .

(3 x 6

(-0 .5 z 1 0 .5328

Q2 .

(10 x 26

(-1 .25 z 2 .75 0 .8719

Q3 .

(50 x 70

(0 .67 z 2 .0 0 .2286

Q .4

(7 x 9

(1 .67 z 3 .33 0 .0471

Q5 .

(8 x 12

(-2 .18 z -0 .94 0 .1590

Q6 .

(40 x 47

(-0 .67 z 30

(z 2 .94 0 .0016

Q8 .

(x 120

(z 1 .33 0 .0918

Q9 .

(x 90

(z -0 .67 0 .7486

Q10 .

(x 2

(z -4 1

Q11 . z -1 .56 Q12 . z -1 .63 Q13 . z 0 .13 Q14 . z 1 .96

Q15 . z 1 .41 Q16 . z 1 .65 Q17 . z -0 .92 Q18 . z -1 .65

Q19 . z 2 .33 Q20 . z 1 .96

Q21 . a )

(x 60

(z -1 0 .8413

b )

(x 60

(z 1 0 .8413

c )

(60 x 110

(-1 z140

(z 2 .2 0 .0139

Q22 . a )

(x 60

(z16

(z -1 .83 0 .9664

c )

(16 x 60

(-1 .83 z60 1-P (x 675

(z 1 .75 0 .0401

b )

(x 450

(z -0 .5 0 .3085

c )

(450 x 675

(-0 .5 z 28

(z 1 .67 0 .0475

e )

(x 12

(z -1 0 .6687

f )

(12 x 18

(-1 z 0 .9 implies z 1 .28

Therefore , SAT x 1 .28 ? or , x 628

ACT x 1 .28 ? or , x 25 .68

b )

(z 0 .8 implies z 0 .84

Therefore , SAT x 0 .84 ? or , x 584

ACT x 0 .84 ? or , x 23 .04

c )

(z 0 .6 implies z 0 .36

Therefore , SAT x 0 .36 ? or , x 536

ACT x 0 .36 ? or , x 20 .16

Q25 . Given 48 8

a ) x 36 implies z -1 .12

Therefore ,

(x 36

(z -1 .12 0 .1314

Therefore , 13 .14 of the batteries can be expected to be replaced by the company

9 11 13 15 17 19 21

6 9 12 15 18 21 24

6 8 10 12 14 16 18

3 6 9 12 15 18 21

x1 1272

1122 x2 ...