1.What proportion of the normal distribution is located between each of the foll
ID: 3207909 • Letter: 1
Question
1.What proportion of the normal distribution is located between each of the following z-score boundaries?
z= -0.50 and z= +0.50
z= -0.90 and z= +0.90
z= -0.90 and z= +0.50
2.Find the z-score boundary that separates a normal distribution as described in each of the following:
The lowest 20% of scores
the top 40% of scores
The bottom 75% of scores
The top 99% of scores
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?
Explanation / Answer
Q1.
Mean ( u ) =0
Standard Deviation ( sd )=1
Normal Distribution = Z= X- u / sd ~ N(0,1)
a.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < -0.5) = (-0.5-0)/1
= -0.5/1 = -0.5
= P ( Z <-0.5) From Standard Normal Table
= 0.30854
P(X < 0.5) = (0.5-0)/1
= 0.5/1 = 0.5
= P ( Z <0.5) From Standard Normal Table
= 0.69146
P(-0.5 < X < 0.5) = 0.69146-0.30854 = 0.3829
38.29% are b/w -0.5 to +0.50
b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < -0.9) = (-0.9-0)/1
= -0.9/1 = -0.9
= P ( Z <-0.9) From Standard Normal Table
= 0.18406
P(X < 0.9) = (0.9-0)/1
= 0.9/1 = 0.9
= P ( Z <0.9) From Standard Normal Table
= 0.81594
P(-0.9 < X < 0.9) = 0.81594-0.18406 = 0.6319
63.19% are b/w -0.9 to +0.90
c.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < -0.9) = (-0.9-0)/1
= -0.9/1 = -0.9
= P ( Z <-0.9) From Standard Normal Table
= 0.18406
P(X < 0.5) = (0.5-0)/1
= 0.5/1 = 0.5
= P ( Z <0.5) From Standard Normal Table
= 0.69146
P(-0.9 < X < 0.5) = 0.69146-0.18406 = 0.5074
50.74% are b/w -0.9 to +0.50
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