An An IQ test was developed for children to have a mean of m

An IQ test was developed for children to have a mean of mu = 75 and a standard deviation of sigma-1 1 What is the probability of randomly picking a child that scored 105 or greater? What is the probability of randomly picking a child that had a scored between 55 and 100. What score will divide the top 45% of the scores from the bottom 55% (round to the nearest whole What is the probability of randomly picking 6 children that had a mean score of 80 or greater? What is the minimum sample size of adults that must be sun eyed in order to be 90% confident the sample percentage is in error by no more than five percentage points assuming that 72% of the respondents believe that the economy is improving?

Solution

Normal Distribution
Mean ( u ) =75
Standard Deviation ( sd )=11
Normal Distribution = Z= X- u / sd ~ N(0,1)                  

1)
P(X < 105) = (105-75)/11
= 30/11= 2.7273
= P ( Z <2.7273) From Standard Normal Table
= 0.9968                  

P(X > = 105) = (1 - P(X < 105)
= 1 - 0.9968 = 0.0032                  

2)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 55) = (55-75)/11
= -20/11 = -1.8182
= P ( Z <-1.8182) From Standard Normal Table
= 0.03452
P(X < 100) = (100-75)/11
= 25/11 = 2.2727
= P ( Z <2.2727) From Standard Normal Table
= 0.98848
P(55 < X < 100) = 0.98848-0.03452 = 0.9540

4)
P(X < 80) = (80-75)/11/ Sqrt ( 6 )
= 5/4.4907= 1.1134
= P ( Z <1.1134) From Standard NOrmal Table
= 0.8672                              
P(X > = 80) = 1 - P(X < 80)
= 1 - 0.8672 = 0.1335              

5)
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.1 is = 1.645
Samle Proportion = 0.72
ME = 0.05
n = ( 1.645 / 0.05 )^2 * 0.72*0.28
= 218.214 ~ 219