IB Maths HL, SL, Studies: July 2013

Monday, July 8, 2013

IB Maths SL Normal Distribution

IB Mathematics SL Normal Distribution
Answers:

1. random variable X normally distributed with mean 25.
The shaded region between 25 and 27 represents 30% of the distribution.
a). Find P(X>27)
Answer:
$P(X>25)=P(25<X<27)+P(X>27)=0.5 \Rightarrow P(X>27)=0.5-P(25<X<27)=0.5-0.3=0.2=20\%$

b). Find the standard deviation of X
Answer:
Let the random variable X , so that

$X\sim N( 25, \sigma)$

We know that $P(X > 27)=0.2$

Since we don’t know the standard deviation of X, we cannot use the inverse normal. Therefore we have to transform the random variable $X$ to that of

$Z\sim N(0,1)$ , using the transformation $Z= \frac{X- \mu}{\sigma}$

we have the following

$P(X > 27)=0.2 \Rightarrow P(\frac{X- 25}{\sigma} > \frac{27- 25}{\sigma})=0.2$

$\Rightarrow P(Z > \frac{2}{\sigma})=0.2$

Using GDC Casio fx-9860G SD
MAIN MENU > STAT>DIST(F5)>NORM(F1)>InvN>

Setting Tail: right
Area: 0.2
$\sigma$ :1
$\mu$ :0

We find that the standardized value is 0.8416

Therefore
$\frac{2}{\sigma}=0.8416 \Rightarrow \sigma =\frac{2}{0.8416}=2.3764$

2. A random variable X is distributed normally with a mean of 20 and variance 9.
a). Find P(X<(or equal to)24.5)
Answer:

By GDC (Casio) menu>STAT>DIST>NORM>Ncd with Lower:-9*10^99, Upper:24.5, $\sigma=9$ , $\mu=20$ and you get
$P(X\leq 24.5)=0.69146246$

b). Let P(X<(or equal to)k)=0.85
ii). Find the value of k.
Answer:

Here you have inverse normal

By GDC (Casio) menu>STAT>DIST>NORM>InvN with Tail:Left, Area:0.85, $\sigma=9$ , $\mu=20$ and you get
$k=29.32$

Question and Answer can be found here

http://www.ibmaths4u.com/viewtopic.php?f=11&t=372

IB Maths HL, SL, Studies

Monday, July 8, 2013

IB Maths SL Normal Distribution

Friday, July 5, 2013

IB Maths Revision Notes HL, SL Studies

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