Answer to Question #201708 in Statistics and Probability for Melinda Burger

Question

1)Assume that the total marks were at an average of u=18 per assesment,and that the distribution of total marks are normally distributed with o=10

a)What is the proportion that a student would have a mark more than 24 mark if randomly selected ?

b)What proportion of students would have marks between 10 and 24

2)A normal distribution has u=80 o=10. What is the probability of random selecting the following scores?

a) x > 75

b) x < 85

c) between the mean and score of 90

d) between the mean and score of 50

e) 75 < x <85

Answer 1

1.)

Given,

$\mu=18\\\sigma=10$

a) Probability that a student have mark more than 24

$P(X>24)$

$z=\dfrac{X-\mu}{\sigma}=\dfrac{24-18}{10}=0.6$

Using Standard normal distribution table:

So, $P(X>24)=P(z>0.6)=0.2743$

b)Probability that a student have marks between 10 and 24

$P(10<X<24)$

$z_1=\dfrac{X-\mu}{\sigma}=\dfrac{10-18}{10}=-0.8\\\ \\z_2=\dfrac{X-\mu}{\sigma}=\dfrac{24-18}{10}=0.6$

Using Standard normal distribution table:

Hence, $P(10<X<24)=P(-0.8<z<0.6)=0.5139$

2.) Given,

$\mu=80\\\sigma=10$

(a) $P(X>75)$

$z=\dfrac{X-\mu}{\sigma}=\dfrac{75-80}{10}=-0.5$

Using Standard normal distribution table:

So, $P(X>75)=P(z>-0.5)=0.6915$

(b) $P(X<85)$

$z=\dfrac{X-\mu}{\sigma}=\dfrac{85-80}{10}=0.5$

Using Standard normal distribution table:

So, $P(X<85)=P(z<0.5)=0.6915$

(c) between the mean and score of 90

$z_1=\dfrac{X-\mu}{\sigma}=\dfrac{80-80}{10}=0$

$z_2=\dfrac{X-\mu}{\sigma}=\dfrac{90-80}{10}=1$

Using Standard normal distribution table:

$P(0<z<1)=0.3413$

(d)  between the mean and score of 50

$z_1=\dfrac{X-\mu}{\sigma}=\dfrac{80-80}{10}=0$

$z_2=\dfrac{X-\mu}{\sigma}=\dfrac{50-80}{10}=-3$

Using Standard normal distribution table:

$P(-3<z<0)=0.4987$

(e) 75 < x <85

$z_1=\dfrac{X-\mu}{\sigma}=\dfrac{75-80}{10}=-0.5$

$z_2=\dfrac{X-\mu}{\sigma}=\dfrac{85-80}{10}=0.5$

Using Standard normal distribution table:

$P(75<X<85)=P(-0.5<z<0.5)=0.3829$