Cumulative Frequency xxxxxxx Table:
Frequency 

1 
1  
30  xxxxxxx nowrap="nowrap" style="width:111px;height:20px;">
9 

40 xxxxxxx 49  8 

10 
27 
60  xxxxxxx nowrap="nowrap" style="width:111px;height:20px;">
39 

70 xxxxxxx 79  7 

2 
45 
90  xxxxxxx nowrap="nowrap" style="width:111px;height:20px;">
47 
(4739)= 17.02%
 The xxxxxxx frequency of xxxxxxx above table xxxxxxx as follows:
Frequency xxxxxxx style="width:95px;height:20px;"> fx 
C.F  
20  xxxxxxx style="width:86px;height:20px;">
19.529.5 
24.5 
1 

30 xxxxxxx 39 
34.5 
8 
276 

39.549.5 
356 
17  
50 xxxxxxx xxxxxxx style="width:86px;height:20px;">
49.559.5 
54.5  10 
27 

60 xxxxxxx 69 
64.5 
12 
774 

69.579.5 
521.5 
46  
80  xxxxxxx xxxxxxx align="center">79.589.5  84.5  2 
90 xxxxxxx 99 
94.5 
2 
189 




57.08
where, xxxxxxx = lower xxxxxxx xxxxxxx the xxxxxxx in which xxxxxxx median lies, h= xxxxxxx interval xxxxxxx xxxxxxx the modal xxxxxxx Frequency of xxxxxxx class interval xxxxxxx which the xxxxxxx lies, cf= Cumulative xxxxxxx of the xxxxxxx xxxxxxx Cumulative xxxxxxx of the xxxxxxx median of xxxxxxx above xxxxxxx xxxxxxx (N/2)^{th}item = xxxxxxx = 25^{th} xxxxxxx median = xxxxxxx value="4">The mean xxxxxxx increase as xxxxxxx increase in xxxxxxx xxxxxxx is xxxxxxx more than xxxxxxx of the xxxxxxx in xxxxxxx xxxxxxx for mean xxxxxxx a grouped xxxxxxx Since the xxxxxxx is having xxxxxxx multiplier effect, xxxxxxx will increase xxxxxxx xxxxxxx times, xxxxxxx the denominator xxxxxxx only increase xxxxxxx 18 xxxxxxx xxxxxxx times. The median xxxxxxx also increase xxxxxxx the median xxxxxxx a grouped xxxxxxx is a xxxxxxx comprising of xxxxxxx xxxxxxx and xxxxxxx is positively xxxxxxx to the xxxxxxx So xxxxxxx xxxxxxx Cumulative frequency xxxxxxx the median xxxxxxx increases.
B = xxxxxxx xxxxxxx the xxxxxxx toss is xxxxxxx of getting xxxxxxx head xxxxxxx xxxxxxx toss of xxxxxxx = ½ Probability xxxxxxx A = xxxxxxx of B= xxxxxxx Probability = xxxxxxx = ¼
= (242+231+220+213+230+293)/6 xxxxxxx 1429/6 = xxxxxxx xxxxxxx = xxxxxxx 22.38
Standard xxxxxxx (s) = xxxxxxx (given) µ) xxxxxxx 95% level xxxxxxx given by xxxxxxx formula:
where, xxxxxxx xxxxxxx 1 xxxxxxx Level/100) = xxxxxxx = 0.05 xxxxxxx this xxxxxxx xxxxxxx = Value xxxxxxx from tdistribution xxxxxxx for the xxxxxxx dfand α/2, Degree of xxxxxxx (df) = xxxxxxx xxxxxxx (n) xxxxxxx 1 = xxxxxxx – 1 xxxxxxx 48 xxxxxxx xxxxxxx sum.
For xxxxxxx interval we xxxxxxx the values xxxxxxx above in xxxxxxx xxxxxxx style="marginleft:35.45pt;">
xxxxxxx style="marginleft:35.45pt;"> = The xxxxxxx xxxxxxx for 95% xxxxxxx level is xxxxxxx 3968.21)

2 
5 
8 

0.1 
0.4 
0.1 
x. 
0.9 
2.0 
1.0 
The xxxxxxx value of xxxxxxx = 0.2+0.9+2.0+0.8+1.0 xxxxxxx xxxxxxx value="19">
5 
10 

0.3 
0.4 
0.1 

0.2  0.9 
0.8 
1.0 
2*0.1 xxxxxxx (34.9)^{2}*0.3 + xxxxxxx + (84.9)^{2}*0.1 xxxxxxx (104.9)^{2}*0.1]
xxxxxxx 0.841+1.083+1.444+0.961+2.601 = xxxxxxx xxxxxxx style="marginleft:35.45pt;">S.D xxxxxxx 2.63
People who xxxxxxx not show xxxxxxx i.e. q xxxxxxx 20% = xxxxxxx bookings of xxxxxxx xxxxxxx = xxxxxxx seat in xxxxxxx flight for xxxxxxx day xxxxxxx xxxxxxx passangers who xxxxxxx up = xxxxxxx * 80% xxxxxxx 9.6 (say xxxxxxx or less)
Hence, xxxxxxx probability of xxxxxxx xxxxxxx given xxxxxxx the formula:
 Binomcdf xxxxxxx Size, Success xxxxxxx Trials)
Hence xxxxxxx xxxxxxx = Binomcdf(12,0.8,10) xxxxxxx 0.725
 Nos xxxxxxx passangers expected xxxxxxx show up xxxxxxx 12*0.8 = xxxxxxx value="22">We use xxxxxxx xxxxxxx E(y) xxxxxxx 726.2 and xxxxxxx = 85.3
z xxxxxxx = xxxxxxx xxxxxxx 2.249
 = 0.10, xxxxxxx Critical z xxxxxxx 0.10 level xxxxxxx 2.000, using xxxxxxx with,
Degree of xxxxxxx = 651=64
As, xxxxxxx xxxxxxx (2.249) xxxxxxx Critical z xxxxxxx hence we xxxxxxx that xxxxxxx xxxxxxx hypothesis can xxxxxxx rejected.
 = xxxxxxx = 1
Taking xxxxxxx from the xxxxxxx xxxxxxx table, xxxxxxx get Z_{1}= xxxxxxx & Z_{2} xxxxxxx 0.34134
Hence xxxxxxx xxxxxxx that a xxxxxxx person has xxxxxxx BestEver credit xxxxxxx between 500 xxxxxxx 700 = xxxxxxx 0.34134 +0.34134 xxxxxxx xxxxxxx value="24">From xxxxxxx z score xxxxxxx we get xxxxxxx the xxxxxxx xxxxxxx for the xxxxxxx percentile is xxxxxxx z = xxxxxxx where x= xxxxxxx of sample xxxxxxx 1.28 = xxxxxxx xxxxxxx value="25">S.D xxxxxxx sample mean xxxxxxx Normal S.D/ xxxxxxx where xxxxxxx xxxxxxx sample observations
Therefore, xxxxxxx of Sample xxxxxxx = 100/ xxxxxxx 10
 Sample xxxxxxx = 524
S.D xxxxxxx 27

In xxxxxxx left tailed xxxxxxx the xxxxxxx xxxxxxx the area xxxxxxx the left xxxxxxx the test xxxxxxx (z= 2.00). xxxxxxx the z xxxxxxx table, the xxxxxxx xxxxxxx the xxxxxxx of 2.00 xxxxxxx 0.228. The xxxxxxx 0.228 xxxxxxx xxxxxxx than 0.01, xxxxxxx null hypothesis xxxxxxx not rejected.

 Female Bosses xxxxxxx harshly critical a= xxxxxxx xxxxxxx align="center">b= 54
The claim xxxxxxx the proportion xxxxxxx women saying xxxxxxx xxxxxxx are xxxxxxx critical is xxxxxxx than the xxxxxxx of xxxxxxx xxxxxxx female bosses xxxxxxx harshly critical xxxxxxx be expressed xxxxxxx p_{2}
1< xxxxxxx then p_{1}≥ xxxxxxx xxxxxxx style="marginleft:54.0pt;">Because xxxxxxx the claim xxxxxxx p_{2} does xxxxxxx contain xxxxxxx xxxxxxx becomes the xxxxxxx hypothesis. The xxxxxxx hypothesis is xxxxxxx statement of xxxxxxx so we xxxxxxx style="liststyletype:upperalpha;">
 , 0.7405
The test xxxxxxx (z) = xxxxxxx xxxxxxx (55/220)(54/200) xxxxxxx 0.467
This being xxxxxxx left xxxxxxx xxxxxxx the Pvalue xxxxxxx the area xxxxxxx the left xxxxxxx the test xxxxxxx style="marginleft:54.0pt;">(z= 0.467). xxxxxxx the z xxxxxxx xxxxxxx the xxxxxxx to the xxxxxxx of 0.467 xxxxxxx 0.3228. xxxxxxx xxxxxxx 0.3228 is xxxxxxx than the xxxxxxx significance value xxxxxxx the null xxxxxxx of p_{1} xxxxxxx p_{2} is xxxxxxx xxxxxxx style="marginleft:54.0pt;">We, xxxxxxx that the xxxxxxx of women xxxxxxx female xxxxxxx xxxxxxx harshly critical xxxxxxx higher than xxxxxxx proportion of xxxxxxx saying female xxxxxxx are harshly xxxxxxx As we xxxxxxx xxxxxxx null xxxxxxx our claim xxxxxxx true and xxxxxxx proved.
From the xxxxxxx data k= xxxxxxx = 51 xxxxxxx 4

E = xxxxxxx = 21
For xxxxxxx xxxxxxx obs.: xxxxxxx (2321)^{2}/21 = xxxxxxx style="marginleft:35.45pt;">For the xxxxxxx obs.: xxxxxxx xxxxxxx 2/21 = xxxxxxx style="marginleft:35.45pt;">For the xxxxxxx obs.: x^{2}= xxxxxxx 2/21 = xxxxxxx style="marginleft:35.45pt;">For the xxxxxxx obs.: x^{2}= xxxxxxx xxxxxxx = xxxxxxx style="marginleft:35.45pt;">For the xxxxxxx obs.: x^{2}= xxxxxxx 2/21 xxxxxxx xxxxxxx style="marginleft:35.45pt;">
Total xxxxxxx X^{2}= 0.857
Given xxxxxxx xxxxxxx = xxxxxxx style="marginleft:35.45pt;">As the xxxxxxx level of xxxxxxx is xxxxxxx xxxxxxx the significance xxxxxxx hence the xxxxxxx that such xxxxxxx and illnesses xxxxxxx with equal xxxxxxx on the xxxxxxx xxxxxxx of xxxxxxx week is xxxxxxx for most xxxxxxx the xxxxxxx xxxxxxx value="29">& 30. xxxxxxx given:


Sums  
0 
– xxxxxxx xxxxxxx align="center">1 
2 
3  
2 
– 2  5 
6 
15  
0 
1 
1 
4 
y^{2} 
4 
25 
16 
85 
xy 
2 
5 
12 
23  
)^{2} 
2.56 
0.16 
1.96 
)^{2} 
25 
4 
9 
40 

8 
0.8 
4.2 
Corelation xxxxxxx (r) xxxxxxx xxxxxxx style="marginleft:35.45pt;"> xxxxxxx 14/(5.2 * xxxxxxx = 0.9707 Slope of xxxxxxx regression line xxxxxxx = xxxxxxx xxxxxxx = xxxxxxx style="marginleft:35.45pt;"> Intercept xxxxxxx the regression xxxxxxx (a) xxxxxxx xxxxxxx = (15/5)2.6923(3/5) xxxxxxx style="marginleft:35.45pt;"> The xxxxxxx equation is xxxxxxx = a+bx xxxxxxx 1.38462+ 2.6923x Predicted y’, xxxxxxx xxxxxxx y’= xxxxxxx = 1.38462+ xxxxxxx 4.07692
