Question

Suppose that IQs of East State University’s students can be described by a Normal model with mean 130 and standard deviation 8 points. Also suppose that IQs of students from West State University can be described by a Normal model with mean 120 and standard deviation 10.

a) We select a student at random from East State. Find the probability that this student’s IQ is at least 125 points.
b) We select a student at random from each school. Find the probability that the East State student’s IQ is at least 5 points higher than the West State student’s IQ.
c) We select 3 West State students at random. Find the probability that this group’s average IQ is at least 125 points.
d) We also select 3 East State students at random. What’s the probability that their average IQ is at least 5 points higher than the average for the 3 West Staters?

Answer 1

Answer:

a) $P(X \geq 125) = 1-P(X<125) = 1-P(Z<\frac{125-130}{8}) = 1-P(Z<-0.625) = 1-0.266= 0.734$

b) $P(R \geq 5) = 1-P(R<5) = 1-P(Z< \frac{5-10}{12.806}) =1-P(Z<-0.3904) = 1-0.348=0.652$

c) $P(\bar Y \geq 125) = P(Z> \frac{125-120}{5.774}) = 1-P(Z<0.866) = 1-0.807= 0.193$

d) $P(H\geq 5)= P(Z> \frac{5-10}{7.394}) = 1-P(Z<-0.676)= 1- 0.250=0.750$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the scores for the East population, and for this case we know the distribution for X is given by:

$X \sim N(130,8)$  

Where $\mu_x=130$ and $\sigma_x=8$

Let Y the random variable that represent the scores for the West population, and for this case we know the distribution for Y is given by:

$X \sim N(120,10)$  

Where $\mu_y=120$ and $\sigma_y=10$

Part a

For this case we want this probability:

$P(X\geq 125)$

And we can use the z score given by:

$z = \frac{X -\mu}{\sigma}$

And if we replace we got:

$P(X \geq 125) = 1-P(X<125) = 1-P(Z<\frac{125-130}{8}) = 1-P(Z<-0.625) = 1-0.266= 0.734$

Part b

For this case we need to define the following random variable R = X-Y and we know that the distribution of R is given by:

$R \sim N (130-120= 10, \sigma_R = \sqrt{8^2 +10^2}=12.806)$

And we want this probability:

$P(R\geq 5)$

We can use the z score given by:

$z= \frac{R -\mu_R}{\sigma_R}$

If we use this formula we got:

$P(R \geq 5) = 1-P(R<5) = 1-P(Z< \frac{5-10}{12.806}) =1-P(Z<-0.3904) = 1-0.348=0.652$

Part c

For this case we select a sample size of n =3 for the Y distribution, the sample mean have the following distribution:

$\bar Y \sim N(120, \frac{10}{\sqrt{3}}=5.774)$

And we want this probability:

$P(\bar Y \geq 125) = P(Z> \frac{125-120}{5.774}) = 1-P(Z<0.866) = 1-0.807= 0.193$

Part d

For this case we define the following random variable $H = \bar X -\bar Y$ and the distribution for H is given by:

$H \sim N (130-120=10, \sigma_H = \sqrt{\frac{8^2 +10^2}{3}}= 7.394)$

And the z score would be given by:

$z = \frac{H -\mu_H}{\sigma_H}$

And if we find the probability required we got:

$P(H\geq 5)= P(Z> \frac{5-10}{7.394}) = 1-P(Z<-0.676)= 1- 0.250=0.750$