1.
Determine whether each of the distributions given below represents a
probability distribution.
Justify your answer.
(A)
x
1
P(x) 3/10
2
1/5
3
1/5
4
3/10
(B)
x
3
P(x) 0.2
6
3/5
8
3/10
(C)
x
20
P(x) 1/10
30
0.1
40
1/2
50
0.3
Answer (A)
𝑃(𝑥) ≥ 0 for all 𝑥.
3
1
1
3
3+2+2+3
10
∑ 𝑃(𝑥) = + + + =
= 10 = 1
10
5
5
10
10
Therefore this is a probability distribution.
Answer (B)
𝑃(𝑥) ≥ 0 for all 𝑥.
3
3
2
6
3
2+6+3
11
∑ 𝑃(𝑥) = 0.2 + + = + + =
=
≠1
5
10
10
5
10
10
10
Since ∑ 𝑃(𝑥) ≠ 1, this is not a probability distribution.
Answer (C)
𝑃(𝑥) ≥ 0 for all 𝑥.
1
1
1
1
5
3
1+1+5+3
10
∑ 𝑃(𝑥) = + 0.1 + + 0.3 = + + + =
= 10 = 1
10
2
10
10
10
10
10
Therefore this is a probability distribution.
2. Consider a binomial distribution with 15 identical trials, and a
probability of success of 0.5
i.
ii.
Find the probability that x = 3 using the binomial tables
Use the normal approximation to find the probability that x = 3.
Show all work.
(Points :4)
(i) 𝑃(𝑋 = 3) = 0.0139.
(From binomial distribution tables, the value for 𝑥 = 3, 𝑛 = 15 , and 𝑝 = 0.5)
(ii)
Mean = 𝐸(𝑋) = 𝑛𝑝 = 15 × 0.5 = 7.5
Standard deviation = √𝑛𝑝(1 − 𝑝) = √15 × 0.5 × 0.5 = √3.75 = 1.9365
The distribution of 𝑋 can be approximated by the normal distribution with mean 7.5 and
standard deviation 1.9365.
𝑃(𝑋 = 3) can be approximated by 𝑃(2.5 ≤ 𝑋 ≤ 3.5) because normal distribution is
continuous and binomial distribution is discrete.
2.5 − 7.5 𝑋 − 7.5 3.5 − 7.5
𝑃(2.5 ≤ 𝑋 ≤ 3.5) = 𝑃 (
≤
≤
)
1.9365
1.9365
1.9365
= 𝑃(−2.582 ≤ 𝑍 ≤ −2.066) (𝑤ℎ𝑒𝑟𝑒 𝑍 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑛𝑜𝑟𝑚𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛)
= 𝑃(𝑍 ≤ −2.066) − 𝑃(𝑍 ≤ −2.582) = 0.0194 − 0.0049 = 0.0145
3. The diameters of oranges in a certain orchard are normally distributed
with a mean of 5.00 inches and a standard deviation of 0.50 inches.
all work.
Show
(A) What percentage of the oranges in this orchard have diameters less
than 6.8 inches?
(B) What percentage of the oranges in this orchard are larger than 4.90
inches?
(C) A random sample of 100 oranges is gathered and the mean diameter
obtained was 4.90. If another sample of 100 is taken, what is the
probability that its sample mean will be greater than 5.10 inches?
(D)
Why is the z-score used in answering (A), (B), and (C)?
(E) Why is the formula for z used in (C) different from that used in (A)
and (B)?
(Points :10)
Let 𝑋 be the diameter of randomly selected orange. Then 𝑋 follows the normal distribution
𝑋−𝜇
𝑋−5.00
with mean 𝜇 = 5.00 and standard deviation 𝜎 = 0.50. Also 𝑍 = 𝜎 = 0.50 follows the
standard normal distribution.
(A)
𝑋 − 𝜇 6.8 − 5.00
𝑃(𝑋 < 6.8) = 𝑃 (
<
) = 𝑃(𝑍 < 3.6) = 0.9998
𝜎
0.50
Percentage of oranges with diameter less than 6.8 inches = 100 × 𝑃(𝑋 < 6.8)
= 100 × 0.9998 = 99.98
(B)
𝑋 − 𝜇 4.90 − 5.00
>
) = 𝑃(𝑍 > −0.2) = 1 − 𝑃(𝑍 ≤ −0.2)
𝜎
0.50
= 1 − 0.4207 = 0.5793
Percentage of oranges with diameter more than 4.90 inches = 100 × 𝑃(𝑋 > 4.90)
= 100 × 0.5793 = 57.93
(C)
𝑃(𝑋 > 4.90) = 𝑃 (
Let 𝑋̅ be the sample mean. Then 𝑋̅ follows the normal distribution with mean 𝜇 = 5.00 and
𝜎
0.50
0.50
standard deviation 𝑛 =
= 10 = 0.05.
𝑋̅−𝜇
𝑍 = 𝜎/
√𝑛
=
𝑋̅ −5.00
0.05
√100
√
follows the standard normal distribution.
5.10 − 5.00
) = 𝑃(𝑍 > 2) = 1 − 𝑃(𝑍 ≤ 2) = 1 − 0.9772
0.05
= 0.0228
𝑃(𝑋̅ > 5.10) = 𝑃 (𝑍 >
(D)
The Z-scores are made use of in (A), (B) and (C) since tables showing the probability of Z
less than any given value is available. If 𝑋 follows normal distribution with mean 𝜇 and
𝑋−𝜇
standard deviation 𝜎, then 𝑍 = 𝜎 follows standard normal distribution.
(E)
If the standard deviation of 𝑋 is 𝜎, then the standard deviation of 𝑋̅ is 𝜎/√𝑛 . That is why the
𝑋̅ −𝜇
formula 𝑍 = 𝜎/
√𝑛
is used in (C) , while the formula 𝑍 =
𝑋−𝜇
𝜎
is used in (A) and (B).
4. Assume that the population of heights of female college students is
approximately normally distributed with mean

of 68 inches and standard
deviation  of 2.75 inches. A random sample of 16 heights is obtained.
Show all work.
(A) Find the proportion of female college students whose height is greater
than 69 inches.
(B)
Find the mean and standard error of the
distribution
(C) Find
(Points :6)
Let 𝑋 denote the height of a randomly selected female student.
Then 𝑋 follows normal distribution with mean 𝜇 = 68 and standard deviation 𝜎 = 2.75.
Sample size = 𝑛 = 16.
(A)
69−68
𝑃(𝑋 > 69) = 𝑃 (𝑍 > 2.75 ) = 𝑃(𝑍 > 0.3636) = 1 − 𝑃(𝑍 ≤ 0.3636) = 1 − 0.6419 =
0.3581=35.81%
(B)
Mean of 𝑋̅ = 𝜇 = 68 𝑖𝑛𝑐ℎ𝑒𝑠.
𝜎
2.75
2.75
Standard error of 𝑋̅ = 𝑛 =
= 4 = 0.6875
√
√16
(C)
69 − 68
) = 𝑃(𝑍 > 1.4545)
0.6875
= 1 − 𝑃(𝑍 ≤ 1.4545) = 1 − 0.9271 = 0.0729
𝑃(𝑋̅ > 69) = 𝑃 (𝑍 >
5. Answer the following questions regarding the normal, standard normal and
binomial distributions.
(A) Why can the normal distribution be used as an approximation to the
binomial distribution?
(B) What conditions must be met in order to use the normal distribution to
approximate the binomial distribution?
(C) What are the advantages of using the standard normal distribution over
the normal distribution?
(Points :6)
(A)
By the De-Moivre Laplace Central Limit Theorem, the binomial distribution tends to the
normal distribution as the parameter n becomes larger and larger. So for sufficiently large
values of n, the binomial distribution can be approximated by the normal distribution.
If 𝑋 follows binomial distribution with parameters n and p, then it can approximated by the
normal distribution with mean 𝜇 = 𝑛𝑝 and standard deviation 𝜎 = √𝑛𝑝(1 − 𝑝).
(B)
The most important condition is that n should be sufficiently large.
If p is neither near to 0, nor near to 1, the approximation will be more accurate.
(C)
The normal distribution represents a family of distributions. For each combination of values
of 𝜇 and 𝜎 we have a member of the family. The standard normal distribution is that member
of the normal family for which 𝜇 = 0 and 𝜎 = 1. The normal distribution with mean 𝜇 and
standard deviation 𝜎 can be transformed to the standard normal distribution by the
𝑋−𝜇
transformation 𝑍 = 𝜎 . Any probability question regarding X can be answered by
transforming it to Z. Tables are available for the probabilities under the standard normal
distribution.
6. Four cards are selected, one at a time from a standard deck of 52 cards.
Let x represent the number of Jacks drawn in a set of 4 cards.
(A) If this experiment is completed without replacement, explain why x is
not a binomial random variable.
(B) If this experiment is completed with replacement, explain why x is a
binomial random variable.
(Points :4)
(A)
Suppose the experiment is completed without replacement.
At the time of first draw, there are 52 cards out of which 4 are Jacks.
So the probability of drawing a Jack is 4/52.
At the time of second draw, there are 51 cards out of which either 3 or 4 are Jacks depending
on the outcome of the first draw.
So the probability of drawing a Jack is 3/51 or 4/51 depending on the outcome of the first
draw.
So the draws are not independent and the probability of getting a Jack is not the same from
trial to trial. That is why the distribution is not binomial.
Here the distribution of X is hyper geometric distribution.
(B)
Suppose the experiment is repeated with replacement.
Then in every draw there are 52 cards and 4 of them are Jcaks.
So the probability of getting Jack in a particular draw is independent of the outcomes of the
previous draws and we have the same probability 4/52 of drawing Jack in each and every
draw. So the distribution of X is binomial with n=4 and p=4/52=1/13.