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The following data represents the pulse rates of ten randomly selected females after stepping
up and down on a 6-inch platform for 3 minutes. Pulse is measured in beats per minute. 136
169 120 128 129 143 115 146 96 86 (1) Compute the sample mean pulse.
(136 + 169 + 120 + 128 + 129 + 143 + 115 + 146 + 96 + 86)/10 = 1268 / 10 = 126.8
(2) Compute the sample median pulse.
Sorting the data gives 128 and 129 in position 5 and 6
The median is (128 + 129)/2 = 128.5
(3) Compute the range.
Largest – smallest = 169-86 = 83
(4) Compute the sample variance.
s^2 = Σ (x - xbar)^2 / (n-1)
s^2 = (136-126.8)^2 + (169-126.8)^2 + (120-126.8)^2 + (128-126.8)^2 + (129-126.8)^2 + (143126.8)^2) + (115-126.8)^2 + (146-126.8)^2 + (96-126.8)^2 + (86-126.8)^2 /9
s^2 = (9.2)^2 + (42.2)^2 + (-6.8)^2 + (1.2)^2 + (2.2)^2 + (16.2)^2) + (-11.8)^2 + (19.2)^2 + (30.8)^2 + (-40.8)^2 /9
s^2 = 84.64+ 1780.84+ 46.24+ 1.44 + 4.84 + 262.44+ 139.24+ 368.64+ 948.64+ 1664.64/9
s^2 = 5301.6/9
s^2 = 589.0667
(5) Compute the sample standard Deviation.
s = sqrt (589.0667)
s =24.27
Given the following data. X 2 3 5 6 6 Y 10 9 8 3 1
(6) Find X
(2 + 3 + 5 + 6 + 6) / 5 = 4.4
(7) Find Sx
x
x-xbar (x-xbar)^2
2
3
5
6
6
-2.4
-1.4
0.6
1.6
1.6
5.76
1.96
0.36
2.56
2.56
Sx = sqrt(S(x - xbar^2)/(n-1))
Sx = sqrt[(5.76 + 1.96 + 0.36 + 2.56 + 2.56)/4] = 1.8166
(8) Find Y
(10 + 9 + 8 + 3 + 1)/5 = 6.2
(9) Find Sy
y
y-ybar (y-ybar)^2
10
9
8
3
1
3.8
2.8
1.8
-3.2
-5.2
14.44
7.84
3.24
10.24
27.04
Sy = sqrt(S(y - ybar^2)/(n-1))
Sy = sqrt[(14.44 + 7.84 + 3.24 + 10.24 + 27.04)/4] = 3.9623
(10) Compute r
r = -25.4/(1.8166*3.9623*(5-1))
r = -0.882
According to the Information Please Almanac. 80% of adult smokers started smoking before
turning 18 years old. Suppose 10 smokers 18 years old or older are randomly selected and the
number of smokers who started smoking before 18 is recorded. Q.1. Find the probability that
exactly 8 of them started smoking before 18 years of age.
Using the tables of the binomial distribution, P(X = 8) = 0.30199
Q.2. Find the probability that at least 8 of them started smoking before 18 years of age.
P(8 <= X <=10) = 0.30199 + 0.26844 + 0.10737 = 0.6778
Q.3. Determine the area under the standard normal curve that lies between Z = -2.55 and Z =
2.55
p (Z < -2.55) = 0.0054
0.9946 - 0.0054 = 0.9892
p(Z < 2.55) = 0.9946
Q.4. Find the Z-score such that the area under the standard normal curve to the left is 0.85.
Using a standard z-table, z = 1.04
Q.5. Find the probability. P(Z<-0.61)
0.2709
Q.6. Find the probability. P(1.23<Z≤ 1.56)
0.9406 - 0.8907 = 0.05
Q.7. Assume the random variable X is normally distributed with mean μ = 50 and standard
deviation σ = 7. Compute P(X ≤ 58)
z = (x - mu)/sigma
z = (58 - 50)/7
z = 8/7
z = 1.1429
p = 0.8735
A 2003 study found that medical student’s mean number of hours worked in a week is 81.7.
Suppose the number of hours worked per week by medical residents is approximately normally
distributed with a standard deviation of 6.9 hours.
Q.8. What is the probability that a randomly selected medical resident works more than 80
hours in a week?
z = (x - mu)/sigma
z = (80 - 81.7)/6.9
z = -1.7/6.9
z = -0.2464
p = 0.4027
BUT since we want "greater than", we need to subtract the probability from 1
p = 1 - 0.4027 = 0.5973
Q.9. What is the probability that a randomly selected medical resident works more than 100
hours in a week?
z = (x - mu)/sigma
z = (100 - 81.7)/6.9
z = 18.3/6.9
z = 2.6522
p = 0.996
BUT since we want "greater than", we need to subtract the probability from 1
p = 1 - 0.996 = 0.004
Q.10. What is the probability that a randomly selected medical resident works less than 60
hours in a week?
z = (x - mu)/sigma
z = (60 - 81.7)/6.9
z = -21.7/6.9
z = -3.1449
p = 0.0008
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