20 points

Suppose that IQs of East State University’s students can be described by a Normal model with mean 130 and standard deviation 8 p

lianna [129]

Answer:

a) $P(X \geq 125) = 1-P(X$

b) $P(R \geq 5) = 1-P(R$

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

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

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:

$P(X \geq 125) = 1-P(X$

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:

$P(R \geq 5) = 1-P(R$

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:

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

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:

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

8 0

3 years ago

Please Answer ASAP! Answer MUST include an explanation in order to receive points and the Brainliest answer. Thank you.

miss Akunina [59]

Probablity is (desired outcomes)/(total possible outcomes)

total possible outcomes is 50

find the desired oucomes
les than 10
those ar
1,2,3,4,5,6,7,8,9
9 numbers less than 10

OR
multipule of 12
12,24,36,48
4 of them

9+4=13
13=total possible

13/50
C

7 0

4 years ago

Read 2 more answers

(problem 6.13 page 97) Consider a population in which 80% of males and 60% of females are employed. In this population, 55% of i

svp [43]

Answer:

There is a 50.77% probability that no more than one of those chosen is not employed.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they will be employed, or they will not. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

We have these following percentages:

55% of the individuals are female.

So 45% of the individuals are male.

80% of males are employed. So 20% of males are unemployed.

60% of females are employed. So 40% of females are unemployed.

If I pick five persons at random from this population, what is the probability that no more than one of those chosen is not employed?

Using the binomial distribution, p is the probability that a person is unemployed. 40% of the females and 20% of the males are unemployed. The population is 55% females and 45% males. So

$p = 0.4*0.55 + 0.2*0.45 = 0.31$

There are five persons, so $n = 5$

What is the probability that no more than one of those chosen is not employed?

$P(X \leq 1) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.31)^{0}.(0.69)^{5} = 0.1564$

$P(X = 1) = C_{5,1}.(0.31)^{1}.(0.69)^{4} = 0.3513$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.1564 + 0.3513 = 0.5077$

There is a 50.77% probability that no more than one of those chosen is not employed.

7 0

3 years ago

What’s The Answer To This? Please Help Fast

s2008m [1.1K]

Answer:

x=30

Step-by-step explanation:

you would call those alternate exterior angles meaning that they are the same angle, so all you have to do is reverse the equation meaning you do 75-15=60 and you would then divide by 2 to get X so 60/2=30, so <u>x=30</u>

4 0

2 years ago

Read 2 more answers

If y is directly proportional to x and y = 17.5 when x=21 find y when x=39

morpeh [17]

direct proportion y=kx , k is constant and k does not equal to 0.

17.5=k(21)

k=5/6

when x =39

y=5/6 ×39

y=32.5

4 0

3 years ago

Read 2 more answers