# Density Distribution + Standardised Normal Distribution

• Aug 25th 2009, 06:55 PM
rel85
Density Distribution + Standardised Normal Distribution
Hi there,

I have 2 questions to ask and would be very appreciative as to learning the methods to solving these type of questions in a descriptive manner. Thanks will be given!

Q1)
Ralph is a great darts player. The probability that he hits the bull's-eye is 2/3. He throws 5 times.

i) First write the density distribution in table form.

Then calculate the probability (with four significant digits) that he hit the bull's-eye:

i) exactly 3 times
ii) at least 3 times
iii) not at all

Q2)
We suppose that for male students at university their weight follows a normal distribution with a mean of 150 pounds and a standard deviation of 25 pounds. Find the percentage of male students whose weight is:

i) In between 125 and 175 pounds
ii) In between 120 and 200 pounds
iii) Over 200 pounds

Provide your answers in percentages to two decimals in the fraction part. Use the standardised normal distribution table...
• Aug 25th 2009, 07:48 PM
pickslides
Quote:

Originally Posted by rel85

Q2)
We suppose that for male students at university their weight follows a normal distribution with a mean of 150 pounds and a standard deviation of 25 pounds. Find the percentage of male students whose weight is:

i) In between 125 and 175 pounds
ii) In between 120 and 200 pounds
iii) Over 200 pounds

Provide your answers in percentages to two decimals in the fraction part. Use the standardised normal distribution table...

You need to convert these questions into z-scores.
$Z= \frac{X-\mu}{\sigma}$

Using

$\mu = 150$
$\sigma = 25$

For i) $125

$P(125

For ii) $120

$P(120

For iii) $x> 200$

$P(x> 200) = P\left(Z>\frac{200-150}{25}\right) = \dots$

Normal distribution - Wikipedia, the free encyclopedia

Z table - Normal Distribution
• Aug 25th 2009, 10:39 PM
matheagle
The first one is a binomial, with n=5 and p=2/3.
This is not a continuous random variable so I woldnt call it a density.

X is the number of bulls eyes..

$P(X=x)={5 \choose x}\biggl({2\over 3}\biggr)^x \biggl({1\over 3}\biggr)^{5-x}$

where x=0,1,...,5.

(i) $P(X=3)={5 \choose 3}\biggl({2\over 3}\biggr)^3 \biggl({1\over 3}\biggr)^2$

(ii) $P(X=3)+P(X=4)+P(X=5)$

$= {5 \choose 3}\biggl({2\over 3}\biggr)^3 \biggl({1\over 3}\biggr)^2+{5 \choose 4}\biggl({2\over 3}\biggr)^4 \biggl({1\over 3}\biggr)^1+{5 \choose 5}\biggl({2\over 3}\biggr)^5 \biggl({1\over 3}\biggr)^0$

(iii) $P(X=0)={5 \choose 0}\biggl({2\over 3}\biggr)^0 \biggl({1\over 3}\biggr)^5$