Answer to Question #124655 in Statistics and Probability for 12kjdklll

Question

The heights of 500 students are normally distributed with a mean of 144.5 centimeters and a variance of 36 centimeters. Assuming that the
heights are recorded to the nearest half-centimeter,
find probability,
(a) less than 130.0 centimeters?
(b) between 151.5 and 182.0 centimeters inclusive?
(c) greater than or equal to 150.0 centimeters?
(d) p(x>x0)=0.6

Answer 1

general

Cumulative distribution function of normal distribution is

$F\left( x \right) = \frac{1}{2}\left(1 + \mathop{\mathrm{erf}}\nolimits \left(\frac{x-\mu}{\sqrt{2}\sigma}\right) \right)$

Here $\mu = 144.5, \sigma = 36$ , so

$F\left( x \right) = \frac{1}{2}\left(1 + \mathop{\mathrm{erf}}\nolimits \left(\frac{x-144.5}{36\sqrt{2}}\right) \right)$,

where erf function is

$\mathop{\mathrm{erf}}\nolimits\left(x\right) = \frac{2}{\sqrt{\pi}} \int\limits_0^x e^{-t^2} dt$ .

To find a value of erf function, we have to check a special table of values or just use a calculator.

a)

Solution:

$P\left(X < 130\right) = F\left( 130 \right) = \frac{1}{2} \left( 1 + \mathop{\mathrm{erf}}\nolimits \left( \frac{-14.5}{36\sqrt{2}} \right) \right) \approx \frac{1}{2} \left( 1 - 0.3129 \right) = 0.34355\\[0.3cm]$

Answer: 0.34355

b)

Solution:

$P\left( 151.5 < X < 182 \right) = F\left( 182 \right) - F\left( 151.5 \right) = \\[0.3cm]$

$\quad = \frac{1}{2} \left( 1 + \mathop{\mathrm{erf}}\nolimits \left( \frac{37.5}{36\sqrt{2}} \right) \right) - \frac{1}{2} \left( 1 + \mathop{\mathrm{erf}}\nolimits \left( \frac{7}{36\sqrt{2}} \right) \right) \approx \\[0.3cm]$

$\quad \approx \frac{1}{2} \left( 1 + 0.7024 \right) - \frac{1}{2} \left( 1 + 0.1542 \right) \approx 0.274 \\[0.3cm]$

Answer: 0.274

c)

Solution:

$P\left( X \geq 150 \right) = 1 - F\left( 150 \right) = 1 - \frac{1}{2}\left( 1 + \mathop{\mathrm{erf}}\nolimits \left( \frac{5.5}{36\sqrt{2}} \right) \right) \approx \\[0.3cm]$

$\quad \approx 1-\frac{1}{2}\left( 1 + 0.1214 \right) \approx 0.4393 \\[0.3cm]$

Answer: 0.4393

d)

Solution:

$P\left( X > x_o \right) = 1 - F\left( x_o \right) = 1 - \frac{1}{2}\left( 1 + \mathop{\mathrm{erf}}\nolimits \left( \frac{x_o - 144.5}{36\sqrt{2}} \right) \right) = 0.6 \\[0.3cm]$

$\Longrightarrow \quad \mathop{\mathrm{erf}}\nolimits \left( \frac{x_o - 144.5}{36\sqrt{2}} \right) = -0.2 \\[0.3cm]$

$\Longrightarrow \quad \frac{x_o - 144.5}{36\sqrt{2}} \approx -0.1791 \\[0.3cm]$

$\Longrightarrow \quad x_o \approx 135.382 \approx 135.5\\[0.3cm]$

Answer: 135.5