Question

Assume the random variable x is normally distributed with mean u = 80 and standard deviation c=5. Find the indicated probability. P(65<x< 73) P(65<x< 73)=0 (Round to four decimal places as needed.)

Answer 1

We have given that X be the random variable normally distributed with mean $\mu$ =80 and standard deviation $\sigma$ =5

X~N(80,5²)

So, to find

P(65<X<73)

73 – 80 65 - 80 = PG 5 X – 80 5 5

$=P(-3<Z<-1.4)$ [Z~N(0,1)]

$=\phi(-1.4)-\phi(-3)=0.0794$

5.2.7

Let X be the random variable denoting mean  $\mu$=507 and standard deviation $\sigma$ =122

To find

a) P(X<550)

$=P(\frac{X-507}{122}<\frac{550-507}{122})$

$=P(Z<0.352459)$

$=\phi(0.352459)$

=0.6378

approximately 63.78% sat score less than 550.

b)

here first find,

P(X>525)

$=P(\frac{X-507}{122}>\frac{525-507}{122})$

=P(Z>0.147541)

$=1-\phi(0.147541)$

=0.4414

n=1000 sample has taken, so 1000*0.4414=441.4 samples are approx greater than 525.

(442 samples in integer)

#########################part c

In this problem X~N(100,13²)

a) P(X<66)

$=P(\frac{X-100}{13}<\frac{66-100}{13})$

=P(Z<-2.615385)

$=\phi(-2.615385)$

=0.0045

b)

P(88<X<110)

$=P(\frac{88-100}{13}<\frac{X-100}{13}<\frac{110-100}{13})$

=P(-0.9230769<Z<0.7692308)

$=\phi(0.7692308)-\phi(-0.9230769)$

=0.6011

c)

P(X>130)

$=P(\frac{X-100}{13}>\frac{130-100}{13})$

=P(Z>2.307692)

$=1-\phi(2.307692)$

=0.0105