1. SOLUTION.—

Let X denotes the scores of 12th- grade students on the National Assessment of Educational Progress.
Mean = 299 and standard deviation = 34.

X ~ N (µ= 299 and σ = 34)

A) The probability that his or her score is higher than 299 P(X > 299) = P (Z > (299-299/34)) [Since Z = (X - µ)/ σ] = P (Z > 0) = 0.5 (Ans.)

B) The probability that his or her score is higher than 367 P(X > 367) = P (Z > (367-299/34)) [Since Z = (X - µ)/ σ] = P (Z > 2) = 0.02275 (Ans.)

C) SRS of 4 , 12-graders..
First take 280,290,300,310 theirs x bar=(280+290+300+310)/4=295
Then do this again…
Take 282,295,320,330 theirs x bar= 306.5
If we take 291,270,300,333 then x bar=298.5

If we take 284,275,304,326 then x bar= 297.25

Now
Mean of all x bar= (295+306.5+298.5+297.25)/4=299.31(Ans.)

D) Standard Deviation of all x bar= 5.13(Ans.)

E) If x=299 then z= (299-299.31)/(5.13/sqrt(4))= -0.12
Now p(x>299)=P(z>-0.12)=0.5+0.0478=0.5478(Ans.)

F) If x=367 then z= (367-299.31)/ (5.13/sqrt(4))= 26.38
Now p(x>367)=P(z>26.38)=0(Ans.)

2. SOLUTION.—

Sample size (n) =50
Population mean (µ) =0.6
Population S.D. (σ) =1

Since sample size is sufficiently larger and σ is known we can apply central limit theorem here.

According to the central limit theorem, the mean of the sampling distribution of means,
The mean of the average number of moths in 50 traps is
E(x-bar) = µ = 0.6 (Ans.)

According to the central limit theorem, the standard error of the mean SE(x-bar)
= σ/ √ n
= 1/ √ 50
= 0.141 (Ans.)

The probability (±0.01) that the average number of moths in 50 traps is greater than 0.7
X-bar =0.7
Z = (x-bar - µ)/ σ/ √ n = (0.7-0.6) / 0.141 = 0.709 P (X-bar>0.7)
P (Z=.709)
= 0.2852
So, the probability is 28.52% (Ans.)

3. SOLUTION.—

a) Mean (µ) =1040
And S.D. (σ) =210
X-bar = 1066
Z = (x-bar - µ)/ σ = (1066 - 1040)/ 210 = 0.12

So, P(X-bar=1066) =P (Z = 0.12) = 0.0478 = 4.78% (Ans.)
3 b.

Test statistic z=(1066-1040)/(210/sqrt(60))=0.96

P(z140)=?

Test statistic, z= (140-116)/14=1.71
P(x>140)=P(z>1.71)=0.5-P(0140)=P(z>2.42)=0.5-(02.33)=0.01

So, (x - 115)/(11.6/√(5))=2.33
Or, x= 2.33*11.6/√(5) +115=127.08

L=127.08(Ans.)