1. solution for midterm exam
Ch 01 – Ch 06

2. 1. Calculate each value requested for the following set of scores (2 points each) X: 1, 3, 0, 2 and Y: 5, 1, –2, –4
a. ΣX = 6
b. ΣY = 0
c. ΣXY = 0
d. ΣX2 = = 14
e. Σ(Y – 2)2 = = 62

3. 2.For the distribution shown in the following table: (3 points each)X f cf c%
25-29 4 25
100
20-24 6 21 84
15-19 7 15 60
10-14 5 8 32
5-9 3 12
Find the percentile rank for X = (32%)
Find the 60th percentile. (19.5)
Find the percentile rank for X = 11. (18%)
Find the 66th percentile. (20.75)

4. 2. c-d c. (11-9.5)/(14.5-9.5) = (x-0.12)/(0.32-0.12)
1.5/5=(x-0.12)/0.2 0.3*0.2=x-0.12
 =x x = 0.18
d. (y-19.5)/( )=( )/( )
(y-19.5)/5=0.06/0.24y-19.5=5*0.25=1.25
y = = 20.75

5. 3. (5 points each)
A set of n = 6 scores has a mean of M = 10. Another set of scores has n = 4 and M = 15. If these two sets of scores are combined, what is the mean for the combined group? (n=10, ΣX = 120, M = 12)
A sample of n = 7 scores has a mean of M = 6. If one score with a value of X = 12 is removed from the sample, what is the mean for the remaining scores? (ΣX = 42 ΣX’ = 42-12=30M’=30/6=5)
A sample of n = 6 scores has a mean of M = 10. One new score is added to the sample and the new mean is computed to be M = 9. What is the value of the score that was added to the sample? (ΣX = 60 ΣX’ =60+x=9*7, X=3)
For a sample with M = 40 and s = 4, the middle 95% of the individuals will have scores between what scores? M2s=(32,48)

6. 4. (5 points each)
A population has µ = 50 and σ = 5. If 10 points are added to every score in the population, then what are the new values for the mean and standard deviation? (µ = and σ = 5)
A population of scores has µ = 50 and σ = 5. If every score in the population is multiplied by 3, then what are the new values for the mean and standard deviation? (µ = 50*3 and σ = 5*3)
Using the definitional formula, compute SS, variance and the standard deviation for the following sample of scores. Scores: 3, 6, 1, 6, 5, 3 (M=4, SS = 20; s 2 = 4; s = 2)

7. 5. (9 points each)
A population of scores with µ = 73 and σ = 20 is standardized to create a new population with µ = 50 and σ = 10. What is the new value for each of the following scores from the original population? Scores: 63, 65, 77, 83
For a distribution of scores, X = 40 corresponds to a z‑score of z = +1.00, and X = 28 corresponds to a z‑score of z = – What are the values for the mean and standard deviation for the distribution? (Hint: Sketch a distribution and locate each of the z‑score positions.)

8. 5. X=63z=(63-73)/20=−0.5(X’-50)/10= − 0.5 X’=45b.
(40 − µ)/ σ = 1  µ + σ = 40
(28 − µ)/ σ = −0.5  µ − 0.5 σ = 28
1.5 σ = = 12  σ =8  µ = 40-8 = 32

9. 6. A normal distribution has a mean of µ = 100 with σ = 20
6. A normal distribution has a mean of µ = 100 with σ = 20. Find the following probabilities: (3 points each)
use z table
a. p(X > 102)=p(z> /20) =p(z>0.1)= =0.4602
b. p(X < 65)=p(z<65-100/20)=p(z<-1.75)= =0.0401
c. p(X < 130)= p(z< /20)=p(z<1.5)= =0.9332
d. p(95 < X < 105)= p(95-100/20<z< /20)=p(-0.25<z<0.25)
=2* =
e. What z-score separates the highest 30% from the rest of the scores? p(0<z<z0)=0.1985≈0.2z0=0.52

10. 7. In an ESP experiment subjects must predict whether a number randomly generated by a computer will be odd or even. (hint: probability of odd or even is the same as the probability of head or tail by tossing a fair coin) (5 points each)
a. What is the probability that a subject would guess exactly 18 correct in a series of 36 trials?
With n = 36 and p = q = 1/2, you may use the normal approximation
with µ = np = 18 and  2 = npq = 9   = 3.
X = 18 has real limits of 17.5 and 18.5 corresponding to z = 0.17 and z =
p(x=18) = p (-0.17<z<0.17)=2*p(0<z<0.17)=2*0.0675= 0.135

11. 7b.
b. What is the probability that a subject would guess more than 20 correct in a series of 36 trials?
X > 20  X > 20.5  z > /3=0.83
p(X>20.5) = p(z>0.83) = 0.5-p(0<z<0.83) = =