3.For a normal distribution with a mean of=80 and a standard deviation of sx=20.

Question

3.For a normal distribution with a mean of=80 and a standard deviation of sx=20. Find the proportion of the population corresponding to each of the following scores. Scores greater than 85 Score less than 100 Scores between 70 and 90

4. IQ test scores are standardized to produce a normal distribution with a mean of =100 and a standard deviation of sx=15. Find the proportion of the distribution in each of the following IQ categories. Genius or near genius: IQ greater than 140 Very superior intelligence: IQ between 120 and 140 Average or normal intelligence: IQ between 90 and 109

5.The distribution of scores on the SAT is approximately normal with a mean of =500 and a standard deviation of sx=100. For the distribution of students who have taken the SAT, What proportion have SAT scores greater than 700? What proportion have SAT scores greater than 550? What is the minimum SAT score needed to be in the highest 10% of the population? If the state college only accepts students from the top 60% of the distribution, what is the minimum SAT score needed to be accepted?

1.Four airplanes from different airlines have crashed in the past two week. This terrifies Megan, who must travel on a plane. Her travel agent claims that the probability of a plane crash is minuscule. Who is correct and why?

Explanation / Answer

Solution:-
3) X n(80; 20)

a) P(X 85) = P((X - 80)/20 (85 - 80)/20) = P(Z 2.5) = 0.0062

b) P(X 100) = P(( - 80)/20 (100 - 80)/20) = P(Z 1) = 0.8413

c) P(70 X 90) = P((70 - 80)/20 (X - 80)/20 (90 - 80)/20) = P(- 0.5 Z 0.5) = P(Z 0.5) - P(Z - 0.5)
= 0.6915 - 0.3085 = 0.383

4)
a) z = (140 – 100) / 15 = 2.67
The areas in the unit normal distribution, find the area above a z = 2.67. That area is the proportion of “genius or near genius” people: 0.0038

b)Very superior intelligence: IQ between 120 and 140
z = (140 – 100) / 15 = 2.67
z = (120 – 100) / 15 = 1.33

The proportion equals the area below z = 2.67 minus the area below z = 1.33.
In a table of areas under the unit normal distribution, find the area below z = 2.67. That value equals 0.9962. In the table, find the area below z = 1.33. That value equals 0.9082. Take the difference: 0.99620 –0.9082 = 0.0880.

c) Average or normal intelligence: IQ between 90 and 109
z = (109 -100) / 15 = 0.60
z = (90 – 100) / 15 = -0.67

The proportion equals the area below z = 0.60 minus the area below z = -0.67.
In a table of areas under the unit normal distribution, find the area below z = 0.60. That value equals 0.7257. In the table, find the area below z = -0.67. If the table does not have that value, use the area above z= 0.67. That value equals 0.2514. Take the difference: 0.7257 – 0.2514 = 0.4743.

5) a) Find P( X > 700 )
P( ( X - ) / > ( 700 - 500 ) / 100 )
= P( Z > 2 )
= P( Z < -2 )
= 0.02275013

b) Find P(X > 550) = 0.50, this is true for any normal random variable, the probability of being on either side f the mean is 50%.

c) We know from the standard normal that
P(Z < z) = 0.9
= P(Z > z) = 0.1
when z = 1.281552

x = + z *
x = 500 + 1.281552 * 100
x = 628.1552

d) P(Z < z) = 0.4
= P(Z > z) = 0.6
when z = -0.2533471

x = + z *
x = 500 + -0.2533471 * 100
x = 474.6653

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.