estimate the number of students having heights i) Less than 62 inches, ii) between 65 and 71 inches.

Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 08

Year : DEC 2015

Given mean $(m) =68$

Standard deviation $(S.D.) =4$

Total no. of students $(n) =500$.

$$z=\dfrac {X-m}{\sigma}=\dfrac{X-68}4$$

Now to evaluate no. of students less than 62 inches i.e. from 0 to 62

When $x=62 ,z=(62-68)/4\\ Z=-6/4 \\ Z=-1.5$

When $x=0, z=(0-68)/4=-17$

$\therefore P(-17 \lt z \lt -1.5) =$ Area from (z=0 to -17)-Area from (z=0 to -1.5)

$=0.499-0.433 \\ =0.0657$$

Note: Evaluating integration on calc.

$\Bigg[\dfrac 1{\sqrt{2\pi}}\int\limits_0^{-1.5}C^{\frac {-1}2z^2}dz\Bigg]$

Probability of students having height less than 62 inches is $0.0657$

$\therefore B=P \times N \\ \therefore B=0.0657 \times 300 $

$\therefore B=33 $students.

2) Now evaluating no. of students having height between 65 and 71 inches.

$\therefore$ when $X=65\\ z=\dfrac {65-68}4\\ z=\dfrac {-3}4 \\ =-0.75$

when $x=71 \\ z=\dfrac {71-68}4=\dfrac 34=0.75$

$ P(-0.75 \lt Z \lt 0.75)= $ Area from (z=0 to z=-0.75)+Area from (z=0 to 0.75)

$\dfrac 1{\sqrt{2\pi}}\int\limits^0_{-0.75}C^{\frac {-1}2z^2}dz + \dfrac 1{\sqrt{2\pi}}\int\limits_0^{0.75}C^{\frac {-1}2z^2}dz \\ =0.2734+0.2734 \\ =0.5468$

Probability of student having height between 65 to 71 inches is $0.5468$

$\therefore $ No. of students $=N \times P \\ =500 \times 0.5468\\ =273.$

No. of students having height between 65 to 71 inches is 273