Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

1.IQs have a normal distribution with µ = 100 and = 15.Organization that only ac

ID: 3385358 • Letter: 1

Question

1.IQs have a normal distribution with µ = 100 and = 15.Organization that only accepts people within the top 2% of IQ. What level of IQ qualifies people for acceptance ?

2.Explain why a distribution with median M = 64, mean µ = 53 and standard deviation = 9 cannot be a normal distribution.

You arrive at a bus stop at 10, knowing that the bus will arrive at some time normally distributed around 10:15 with a standard deviation of 6 minutes

What is the probability that you have to wait longer than 10 minutes?

  If at 10:15 the bus has not yet arrived, what is the probability that you will have to wait at least an additional 10 minutes?   

5.normal distribution with a mean of 266 days and a standard deviation of 16 days,The longest 20% of pregnancies last longer than how many days?

6.  Standard US IQ tests have a mean of 100 and a standard deviation of 15.What percent of the population has an IQ between 100 and 130?

Explanation / Answer

Q1.
Normal Distribution
Mean ( u ) =100
Standard Deviation ( sd )=15
Normal Distribution = Z= X- u / sd ~ N(0,1)                  

P ( Z > x ) = 0.02
Value of z to the cumulative probability of 0.02 from normal table is 2.05
P( x-u/ (s.d) > x - 100/15) = 0.02
That is, ( x - 100/15) = 2.05
--> x = 2.05 * 15+100 = 130.81                  
level of IQ qualifies people for acceptance is 130.81

Q5.
Mean ( u ) =266
Standard Deviation ( sd )=16
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P ( Z > x ) = 0.2
Value of z to the cumulative probability of 0.2 from normal table is 0.84
P( x-u/ (s.d) > x - 266/16) = 0.2
That is, ( x - 266/16) = 0.84
--> x = 0.84 * 16+266 = 279.472                  
The longest 20% of pregnancies last longer than 279.5 days

Q6.
Mean ( u ) =100
Standard Deviation ( sd )=15
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 100) = (100-100)/15
= 0/15 = 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(X < 130) = (130-100)/15
= 30/15 = 2
= P ( Z <2) From Standard Normal Table
= 0.97725
P(100 < X < 130) = 0.97725-0.5 = 0.4772