# Thread: Density Distribution + Standardised Normal Distribution

1. ## 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...

2. 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.
$\displaystyle Z= \frac{X-\mu}{\sigma}$

Using

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

For i) $\displaystyle 125 <x< 175$

$\displaystyle P(125 <x< 175) = P\left(Z<\frac{175-150}{25}\right) - P\left(Z<\frac{125-150}{25}\right) = \dots$

For ii) $\displaystyle 120 <x< 200$

$\displaystyle P(120 <x< 200) = P\left(Z<\frac{200-150}{25}\right) - P\left(Z<\frac{120-150}{25}\right) = \dots$

For iii) $\displaystyle x> 200$

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

Normal distribution - Wikipedia, the free encyclopedia

Z table - Normal Distribution

3. 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..

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

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

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

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

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

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