4. The national average for the new SAT is 1500 (-1500) and the standard deviati

Question

4. The national average for the new SAT is 1500 (-1500) and the standard deviation is 150 ( = 100). Find the Percentile for a score of 1539 for: A single student (n=1) b. a. For a mean of 60 students (n-60) c. For a mean of 530 students (n-530) A national survey reported that the average number of students per teacher in public schools is 15.9 (X = 15.9). The sample size was 1200 (n-1200). Find a point estimate for the population mean. Find the 97% confidence interval of the true mean. Assume the standard deviation was 2.1 students ( 2.1). 5.

Explanation / Answer

4. We have to find the Z value for this question

Z= (X'-mu)/(Sd / sqrt(n))

a.)

X'=1539

mu=1500

Sd=150

n=1

So., Z=(1539-1500)/(150/sqrt(1)) =0.26

and from the z table we ger P(Z=0.26)=0.6025681 (i.e. 60.26%tile)

b.) using the same equation by just changing n to 60 we get

z= (1539-1500)/(150/sqrt(30))=1.424079

and from the z table we ger P(Z=1.424079)=0.9227882 (i.e. 92.28 %tile)

c.)sing the same equation by just changing n to 530 we get

z= (1539-1500)/(150/sqrt(530)) = 5.98565

and from the z table we ger P(Z=5.98565 )=1(approx) (i.e. 100%tile)

5) So at 97% confidence interval we have 1.5% each left from left and right side of the normal distribution curve

and so we have z value of =-2.17009 lower bound and

z value of =2.17009 upper bound

now we have mu=15.9

sd=2.1 and n=1200

so., Z=(X'-mu)/(Sd / sqrt(n)) same equation

-2.17009 = (X'-15.9)/(2.1/sqrt(1200)) & 2.17009 = (X'-15.9)/(2.1/sqrt(1200))

and so solving these equations we get X'= 15.76845 and 16.03155

So for 97% conf interval we should have true mean between  15.76845 and 16.03155.

Hope the above ans has really helped you. Pls upvote the ans if it has helped. Good Luck!!

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