Probability and Statistics Lecture 16

Continued

Probability and Statistics Lecture 15

Normal Distribution

Questions

In an exam we can buy $800$ students. The $average$ and $standarddeviation$ of marks obtained are $40\%$ and $10\%$. Find

1. number of students who will pass if $50\%$ is kept as minimum
2. What should be the minimum score if $350$ Candidates are to be declared passed.
3. How many candidates have scored marks above $60\%$

Answer

Random Variable : its what the average and standard deviation is describing about $X\sim N(\mu =40,\sigma =10)$

1. number of students who will pass if $50\%$ is kept as minimum

$P(X>50)$ is what we are trying to find here

$$P(X>50)=P(\frac{X-M}{\sigma }>\frac{50-40}{10})$$ $$P(X>50)=P(Z>1)$$ $$P(X>50)=0.1557$$ $$\text{no of students who score atleast 50\%}=P(X>50)\times 800$$ $$0.1557\times 800=127$$

2. What should be minimum score if 350 candidates are to be declared pass

We just discovered that the number of students that will pass is 127 by multiplying the probability with 800

$$350=P(X>X_{min})\times 800$$ $$P(X>X_{min})=\frac{350}{800}$$ $$P(X>X_{min})=0.4375$$ $$P(X<X_{min})=0.5-0.4375$$

We have flipped the

Question 2

The mean height of $500$ students is $151cms$ and the $standarddeviation$ is $15cms$ assuming that the heights are normally distributed find

1. The number of students whose height lies between $120$ and $155$ cms

Answer

Random Variable: The random variable here is height of the students they are rather short.

We have been asked this $P(120<X<155)$

$$P(120<X<155)=P\left(\frac{120-155}{15}<\frac{X-\mu }{\sigma }<\frac{155-151}{15}\right)$$ $$P(120<X<155)=P\left(\frac{-31}{2}<Z<\frac{3}{15}\right)$$ $$P(120<X<155)=P\left(-2.07<Z<0.27\right)$$ $$P(120<X<155)=P\left(-2.07<Z)+P(Z<0.27\right)$$ $$P(120<X<155)=P\left(2.07>Z)+P(Z<0.27\right)$$ $$\text{Multiplied by left inequality by -1.}$$ $$P(120<X<155)=0.4808+0.1064$$ $$P(120<X<155)=0.5872$$ $$\text{Multiply that with 500}$$ P(120<X<155)\times 500= 0.5872=293.6$$ Number of students that have height greater than $120cms$ and less than $155cms$ is 293. ( Absolute Value )

Question 3

Fit a Binomial Distribution for the following

X	0	1	2	3	4	5	6	Total
$f$	5	18	28	12	7	6	4	80
Here n is equal to 6 and we can calculate $mean$ from the table.

• $n=6$
• $\text{mean of x}=E(x)=\sum \frac{xf}{f}$
• $mean$ = $2.4$
• $p=2.4$

Plug this in the calculator for $n=6$ and $p=2.4$

Question 4

References

Information

• date: 2025.02.19
• time: 14:08

Continued

Probability and Statistics Lecture 17