Answer to Question #164508 in Statistics and Probability for Nwabisa

Question

Waiting time_(0;3) Number of patients_210

WT_(3;6) NOP_130

WT_(6;9)NOP_75

WT_(9;12)NOP_40

WT_(12;15)NOP_20

WT_(15;18)NOP_15

WT_(18;21)NOP_10

a.calculate the arithmetic mean

b.calculate the variance

c.calculate the coefficient of variation

d.calculate the median

e.calculate the mode

f.calculate the third quartile

h.calculate the fortieth percentile

Answer 1

$x[WT]\ \ \ f[NOP] \ \ x'=\frac{x_{i-1}+x_{i}}{2}$

$(0;3)\qquad210\qquad1.5$

$(3;6)\qquad130\qquad4.5$

$(6;9)\qquad75\qquad\ \ 7.5$

$(9;12)\quad\ \ 40\qquad\ \ 10.5$

$(12;15)\quad 20\qquad\ \ 13.5$

$(15;18)\quad 15\qquad\ \ 16.5$

$(18;21)\quad 10\qquad\ \ 18.5$

$\Sigma{f_i}=210+130+75+40+20+15+10=500$

$\text{Accumulated frequency}, S:$

$210\ 340\ \ 415\ 455\ \ 475\ 490\ 500$

$a)\bar{x}=\frac{\Sigma{x_i'f_i}}{\Sigma{f_i}}$

$\bar{x}=\frac{210*1.5+130*4.5+75*7.5+40*10.5+20*13.5+15*16.5+10*18.5}{210+130+75+40+20+15+10}=5.19$

$b)D=\frac{\Sigma{f_i}(x'_{i}-\bar{x})^2}{\Sigma{f_i}}$

$\bar{x}=\frac{210*(1.5-5.17)^2+130*(4.5-5.17)^2+75*(7.5-5.17)^2}{500}+\newline +\frac{40*(10.5-5.17)^2+20*(13.5-5.17)^2+15*(16.5-5.17)^2+10*(18.5-5.17)^2}{500}=19.04$

$c)\bar{V}=\frac{\sqrt{D}}{\bar{x}}*100\%=\frac{\sqrt{19.5}}{5.19}*100\%=85\%$

$d)\text{The median is the interval (3;6) hence}$

$S_{(3;6)}=340>\Sigma{f_i}$

$M_e= x_0+\frac{h}{f_{me}}(\Sigma\frac{f_i}{2}-S_{me-1})=3+\frac{3}{130}(\frac{500}{2}-210)=3.92$

$e)\text{interval 0, this interval contains the largest number of features}$

$M_0=x_0+h\frac{f_2-f_1}{(f2-f_1)-(f_3-f_2)}=0+3\frac{210-0}{(210-0)-(210-130)}=2.17$

$f)Q_3=x_{Q_3}+h_{Q_3}\frac{0.75\Sigma{f_i-S_{Q_3-1}}}{f_{Q_3}}=6+3*\frac{0.75*500-340}{75}=7.4$

$h)index = 7*0.4 =2.8\approx3$

$\text{fortieth percentile }= 20$