Question

of the random variable x. Suppose x is a normally distributed random variable with u = 34 and 0 = 4. Find a value a. P(x 2 Xo)=5 b. P(X<Xo) = .025 c. P(x>x) = 10 d. P(x > Xo) = .95 Click here to view a table of areas under the standardized normal curve. a. Xo = (Round to the nearest hundredth as needed.)

Answer 1

Given that X is a normally distributed random variable where,

X - Normal (y = 34,0% = 44)

Before we go on to solve the problems let us know a bit about Normal Distribution.

Normal Distribution

A continuous random variable X is said to have a normal distribution if its PDF(Probability Density Function) is given by

fxle) **),ch

its CDF(Cumulative Distribution Function) is given by,

| | | P(X <) = F(t) = 1 fx(t)dt = | -|(G) dt 1-00 J-2012

Notation:

X Normal(u.o

Standard Normal Distribution

A continuous random variable X is said to have a standard normal distribution if its PDF(Probability Density Function) is given by

its CDF(Cumulative Distribution Function) is given by,

P(X <) = (2) = fx(t)dt = -e #² at J- 0 27T

Exact evaluation of ?(x) is not possible but numerical method can be applied. The values of ?(x) has been tabulated extensively in Biometrika Volume I.

Notation:

X ~ Normal 0,1)

Note:

X Normal(u.o

=Z=- ~Normal(0,1)

Coming back to our problem

Here,

X - Normal (y = 34,0% = 44)

For all the sub parts we need to find the value x₀

(a)

PX > 10) = 0.5

=P(*=4> 40 – 4) = 0,5

10 - PZ > = 0.5[Z ~ Normal(0,1)]

»1-P(< ༠ = 4) - 0.5

$\Rightarrow 1-\Phi\left ( \frac{x_{0}-\mu}{\sigma} \right )=0.5$

= ("")=1-05

$\Rightarrow \Phi\left ( \frac{x_{0}-\mu}{\sigma} \right )=0.5$

$[\Phi(0)=0.5\emph{ (From Normal Tables})]$

一 (0*) 40

$\Rightarrow \frac{x_{0}-\mu}{\sigma}=0$

$\Rightarrow x_{0}-\mu=0$

$\Rightarrow x_{0}=\mu$

$[\mu=34,\sigma=4]$

$\Rightarrow x_{0}=34$

(b)

$\Rightarrow P\left ( \frac{X-\mu}{\sigma}< \frac{x_{0}-\mu}{\sigma} \right )=0.025$

$\Rightarrow P\left ( Z< \frac{x_{0}-\mu}{\sigma} \right )=0.025[Z\sim Normal(0,1)]$

一( ) 2005

(2) =1-0(-1)

»1-6-20 = ༩) =0025

$\Rightarrow \Phi\left ( -\frac{x_{0}-\mu}{\sigma} \right )=1-0.025$

$\Rightarrow \Phi\left ( -\frac{x_{0}-\mu}{\sigma} \right )=0.975$

$[\Phi(1.96)=0.975\emph{ (From Normal Tables})]$

$\Rightarrow \Phi\left ( -\frac{x_{0}-\mu}{\sigma} \right )=\Phi(1.96)$

$\Rightarrow -\frac{x_{0}-\mu}{\sigma}=1.96$

$\Rightarrow x_{0}=\mu-(1.96\sigma)$

$[\mu=34,\sigma=4]$

$\Rightarrow x_{0}=34-(1.96*4)$

$\Rightarrow x_{0}=26.16$

(c)

$\Rightarrow P\left ( \frac{X-\mu}{\sigma}> \frac{x_{0}-\mu}{\sigma} \right )=0.10$

Plz> 11 - Ora ) = 0.10[Z ~ Normal(0,1)]

$\Rightarrow 1-P\left ( Z\leq \frac{x_{0}-\mu}{\sigma} \right )=0.10$

$\Rightarrow 1-\Phi\left ( \frac{x_{0}-\mu}{\sigma} \right )=0.10$

$\Rightarrow \Phi\left ( \frac{x_{0}-\mu}{\sigma} \right )=1-0.10$

$\Rightarrow \Phi\left ( \frac{x_{0}-\mu}{\sigma} \right )=0.90$

$[\Phi(1.28)=0.90\emph{ (From Normal Tables})]$

$\Rightarrow \Phi\left ( \frac{x_{0}-\mu}{\sigma} \right )=\Phi(1.28)$

$\Rightarrow \frac{x_{0}-\mu}{\sigma}=1.28$

= πο = μ + (1.28σ)

$[\mu=34,\sigma=4]$

$\Rightarrow x_{0}=34+(1.28*4)$

$\Rightarrow x_{0}=39.12$​​​​​​​

(d)

$\Rightarrow P\left ( \frac{X-\mu}{\sigma}> \frac{x_{0}-\mu}{\sigma} \right )=0.95$

$\Rightarrow P\left ( Z> \frac{x_{0}-\mu}{\sigma} \right )=0.95[Z\sim Normal(0,1)]$

= 1-P(2310=4) = 0,5

71-4(*=“) = 0.95

Φ-) = 1 – Φ(α)

$\Rightarrow \Phi\left ( -\frac{x_{0}-\mu}{\sigma} \right )=0.95$

$[\Phi(1.65)=0.95\emph{ (From Normal Tables})]$

$\Rightarrow \Phi\left ( -\frac{x_{0}-\mu}{\sigma} \right )=\Phi(1.65)$

$\Rightarrow -\frac{x_{0}-\mu}{\sigma}=1.65$

$\Rightarrow x_{0}=\mu-(1.65\sigma)$

$[\mu=34,\sigma=4]$

$\Rightarrow x_{0}=34-(1.65*4)$

$\Rightarrow x_{0}=27.4$​​​​​​​