BBA – 2 (Spring – 2022)
Statistical Inference
Normal Probability Distributions
Assignment #04
Ahmad Jalil Ansari
Contact: obxml.edu@gmail.com
Question: #01
If X is ~N(𝟑𝟎 , 𝟓) find a, b, c and d in
following questions:
i. P(a < X < 40) = 76.54%
ii. P(26 < X < b) = 76.54%
iii. P(X > c) = 0.135%
iv. P(X < d) = 21.19%
𝜇=0
Question: #01 (i)
If X is ~N(𝟑𝟎 , 𝟓) find a, b, c and d in
following questions:
i. P(a < X < 40) = 76.54%
Part(i)
⟹
⟹
⟹
⟹
⟹
⟹
⟹
⟹
𝜇=0
P(a < X < 40) = 0.7654
z1 = (a – 30) / 5
P((a – 30) / 5 < Z < 2.0) = 0.7654
z2 = (40 - 30) / 5 = 2.0
P(Z < 2.0) - P(Z< (a – 30) / 5) = 0.7654
0.97725 - P(Z < (a – 30) / 5) = 0.7654
P(Z < (a – 30) / 5) = 0.97725 - 0.7654
P(Z < (a – 30) / 5) = 0.21186
(a – 30) / 5 = -0.8 [from Cum. Std. Norm. Dist. Table]
P(-0.8) = 0.21186
a – 30 = -4.0
by searching for z-value that relates to
a = 26.0
probability = 0.21186
Question: #01 (ii)
If X is ~N(𝟑𝟎 , 𝟓) find a, b, c and d in
following questions:
ii. P(26 < X < b) = 76.54%
𝜇=0
Part(ii)
P(26 < X < b) = 76.54%
z1 = (26 – 30) / 5 = -0.8
⟹ P((b – 30) / 5 < Z < 2.0) = 0.7654
z = (b - 30) / 5
⟹ P(Z < (b - 30) / 5) - P(Z < -0.08) = 0.7654 2
⟹ P(Z < (b - 30) / 5) – 0.21186 = 0.7654
⟹ P(Z < (b – 30) / 5) = 0.7654 + 0.21186
⟹ P(Z < (b – 30) / 5) = 0.97726
⟹ (b – 30) / 5 = 2.0
[from Cum. Std. Norm. Dist. Table]
P(2.0) = 0.97726
⟹ b – 30 = 10.0
by searching for z-value that relates to
⟹ b = 40.0
probability = 0.97726
Question: #01 (iii & iv)
If X is ~N(𝟑𝟎 , 𝟓) find a, b, c and d in
following questions:
iii. P(X > c) = 0.13.5%
iv. P(X < d) = 21.19%
𝜇=0
Part(iii)
z1 = (c - 30) / 5
P(X > c) = 0.00135
⟹ P(X < c) = 1 - 0.00135
⟹ P(Z < (c - 30) / 5) = 0.99865
⟹ (c - 30) / 5 = 3.0 [from Cum. Std. Norm. Dist. Table]
P(3.0) = 0.99865
⟹ c = 45
by searching for z-value that relates to probability = 0.99865
Part(iv)
P(X < d) = 0.2119
z1 = (d - 30) / 5
⟹ P(X < (d - 30) / 5) = 0.2119
⟹ (d - 30) / 5 = -0.8 [from Cum. Std. Norm. Dist. Table]
⟹ d = 26
P(-0.8) = 0.2119
by searching for z-value that relates to probability = 0.2119
Question: #02
Find the indicated quantities given that X is ~N(80 , 4).
A) Find the value of b such that P(X ≤ b) = 0.9719.
B) Find the value of c such that P(X ≥ c) = 0.9678.
C) Find the values of A & B such that P(A ≤ X ≤ B) = 0.2961, if A and B
are symmetric about mean.
𝜇=0
Question: #02 (A & B)
X ~N(80 , 4).
A) Find the value of b such that P(X ≤ b) = 0.9719.
B) Find the value of c such that P(X ≥ c) = 0.9678.
𝜇=0
Part(A)
z1 = (b - 80) / 4
P(X ≤ b) = 0.9719
⟹ P(X < (b - 80) / 4) = 0.9719
⟹ (b - 80) / 4 = 1.91
⟹ b = 80 + 4(1.91) P(1.91) = 0.9719
by searching for z-value that relates to probability = 0.9719
⟹ b = 87.64
Part(B)
P(X > c) = 0.9678
z1 = (c - 80) / 4
⟹ P(X < c) = 1 - 0.9678
⟹ P(Z < (c - 80) / 4) = 0.0322
⟹ (c - 80) / 4 = -0.46
⟹ c = 80 + 4 (-0.46) P(-0.46) = 0.0322
by searching for z-value that relates to probability = 0.0322
Question: #02 (C)
X ~N(80 , 4).
C) Find the values of A & B such that
P(A ≤ X ≤ B) = 0.2961, if A and B are
symmetric about mean.
𝜇=0
.
Part(C)
z1 = (A - 80) / 4 and z1 = (B - 80) / 4
P(A ≤ X ≤ B) = 0.2961
⟹ P(X < (B - 80) / 4) - P(X < (A - 80) / 4) = 0.2961
Now A and B are symmetrical to center( i.e. mean) therefore
P(X < (B - 80) / 4) = 0.5 + (0.2961/2) = 0.64805 and
P(X < (A - 80) / 4) = 0.5 - (0.2961/2) = 0.35195
P(X < (B - 80) / 4) = 0.64805
As P(Z < 0.38) = 0.64805
⟹ (B - 80) / 4) = 0.38
⟹ B = 80 + 4(0.38)
⟹ B = 81.52
P(X < (A - 80) / 4) = 0.35195
As P(Z < -0.38) = 0.35195
⟹ (A - 80) / 4) = -0.38
⟹ A = 80 + 4(-0.38)
⟹ A = 78.48
Question: #03
Most graduate schools of business require applicants for admission to take
the GMAT exam. Scores on the GMAT are roughly normally distributed with
a mean of 527 and a standard deviation of 112. i) What is the probability of
an individual scoring above 500 on the GMAT? Ii) How high must an
individual score on the GMAT in order to score in the highest 5%?
Part(i)
P(X > 500) = 1 - P(X < 500)
z = (500 – 527)/112 = -0.241
= P(z < -0.241) = 0.40517 = 40.517%
Part(ii)
P(X > a) = 0.05
P(X < a) =0.95
⟹ P(Z< (a – 527) / 112) = 0.95
⟹ (a – 527) / 112 = 1.645
⟹ a = 527 + 112 (1.645)
⟹ a = 711.24
z = (a – 527)/112
Question: #04
The average number of acres burned by forest and range fires in a large
New Mexico county is 4,300 acres per year, with a standard deviation of 750
acres. The distribution of the number of acres burned is normal. i) What is
the probability that between 2,500 and 4,200 acres will be burned in any
given year? Ii) What number of burnt acres corresponds to the 38th
percentile?
X ~N(4300 , 750)
Part(i)
P(2500< X > 4200) = ?
z1 = (2500 – 4300)/750 = -2.4 z2 = (4200 – 4300)/750 = -1.3333
P(2500< X < 4200) = P(X < 4200) – P(X < 2500)
= P(Z < (--1.3333) – P(Z <-2.4) = 0.4483 -.0082 = 0.4401 = 44.01%
Part(ii)
P(X < a) = 0.38 (As P38 means 38% are below a) z = (a – 4300)/750
⟹ P(Z< (a – 4300) / 750) = 0.38
⟹ (a – 4300) / 750 = -0.31
⟹ a = 4300 + 750 (-0.31)
⟹ a = 4067.5
Question: #05
A Theater chain has studied its movie customers to determine how much
money they spend on concessions. The study revealed that the spending
distribution is normally distributed with a mean of $4.11 and a standard
deviation of $1.37. What percentage of customers will spend less than
$3.00 on concessions? What spending amount corresponds to the top 87th
percentile?
X ~N(4.11 , 1.37)
Part(i)
P(X < 3) = ?
z1 = (3 – 4.11)/1.37 = -0.81
P(X < 3) = P(Z < (-0.81)) = 0.20897 = 20.9%
Part(ii)
z = (a – 4.11)/1.37
P(X > a) = 0.87 (As top P87 means 87% are higher than a)
⟹ P(X < a) = 0.13
⟹ P(Z< (a – 4.11)/1.37) = 0.13
⟹ (a – 4.11) / 1.37 = -1.13
⟹ a = 4.11 + 1.37 (-1.13)
⟹ a = $2.56
Question: #06
A machine produces bolts which are ~N(4 , 0.3), where measurements are
in mm. Bolts are measured accurately and any which are smaller than
3.5mm or bigger than 4.4mm are rejected. Out of batch of 500 bolts how
many would be accepted?
X ~N(4 , 0.3)
P(3.5 < X < 3) = ?
Here
z1 = (3.5 – 4) / 0.3 = -1.67
z2 = (4.4 – 4) / 0.3 = 1.33
Now
P(3.5 < X < 3) = P(-1.67 < Z < 1.33)
= 0.9024 - .04746
= 0.86078
Therefore,
The number of acceptable bolts = 500 x 0.86078
= 430.39 ≈ 430 bolts
Question: #07
IQ Tests are measured on a scale which is ~N(𝟏𝟎𝟎 , 𝟏𝟓). A person wants to
form an “Egghead Society” which only admits people with top 1% of IQ
score. What should he has to set as a cut-off point in the test to allow this
happen?
X ~N(100 , 15)
⟹
⟹
⟹
⟹
⟹
P(X > a) = 0.01
P(X < a) = 0.99
P(Z < (a – 100) / 15)) = 0.99
(a – 100) / 15 = 2.3263
a = 100 + 2.3263(15) = 134.8945
a ≈ 135
Question: #08
A manufacturer does not know mean and SD of the diameters of bearing he
is producing. However, a sieving system rejects all ball bearings larger than
2.4 cm and under 1.8 cm in diameter. Out of 1000 ball bearings 8% were
rejected as too small and 5.5% as too big. What is the mean and SD of the
ball bearing produced?
X ~N(𝜇 , 𝜎)
P(1.8 < X) = 0.08
⟹
⟹
⟹
P(X > 2.4) = 0.055
⟹
P(X < 2.4) = 0.945
P(Z < (1.8 – 𝜇) / 𝜎)) = 0.08 ⟹
P(Z < (2.4 – 𝜇) / 𝜎)) = 0.945
(1.8 – 𝜇) / 𝜎 = -1.4
⟹
(2.4 – 𝜇) / 𝜎 = 1.6
𝜇 – 1.4𝜎 = 1.8
⟹
𝜇 + 1.6𝜎 = 2.4
𝜇 – 1.4𝜎 = 1.8
𝜇 + 1.6𝜎 = 2.4
-3.0 𝜎 = -0.6 (Subtracting)
⟹ 𝜎 = 0.2
⟹ 𝜇 = 2.8 (by substituting 𝜎 = 0.2 in any of two equations)
Question: #09
U.S. Air has a maintenance plan that will discontinue using old aircraft,
specifically those that are in the top 5% of the ages of the aircraft. Of the
current aircraft, the mean age is 13.41 years and the SD is known to be 5.2
years. Find the cut-off age for the top 5% of aircraft.
X ~N(13.41 , 5.2)
⟹
⟹
⟹
⟹
⟹
P(X > a) = 0.05
P(X < a) = 0.95
P(Z < (a – 13.41) / 5.2)) = 0.95
(a – 13.41) / 5.2 = 1.645
a = 13.41 + 1.645(5.2) = 21.964
a ≈ 22 years
Question: #10
A manufacturer, producing nails found that the mean and SD of its length is
2.08 and 0.2 respectively. He rejects all nail larger than 2.4 cm and under
1.8 cm in length. From a bunch of nails, 80 were rejected as too small and
55 as too big. How many nails were there in the bunch?
X ~N(2.08 , 0.2)
Too small ⟹ P(X < 1.8)
= P(Z < (1.8 –2.08) / 0.2)
= P(Z < –1.4)
= 0.8076
Let n1 be the number of nails in
the batch
⟹ 𝐶𝑜𝑢𝑛𝑡 𝑓𝑜𝑟 𝑟𝑒𝑗𝑒𝑐𝑡𝑒𝑑 being too
small = 𝑛1 𝑥 0.8076
80
⟹
𝑛1 =
≈ 100
0.8076
Too big ⟹ P(X > 2.4)
= 1- P(Z < (2.4 –2.08) / 0.2)
= 1 - P(Z < 1.6)
= 1 – 09452 = 0.0548
Let n2 be the number of nails in
the batch
⟹ 𝐶𝑜𝑢𝑛𝑡 𝑓𝑜𝑟 𝑟𝑒𝑗𝑒𝑐𝑡𝑒𝑑 being too
big
= 𝑛2 𝑥 0.548
55
⟹
𝑛1 =
≈ 100
0.0548
Thank you
Ahmad Jalil Ansari
Contact: obxml@yahoo.com
Question: #11
A machine produces bolts which are ~N(4 , 0.3), where measurements are
in mm. Bolts are measured accurately and any which are smaller than
3.5mm or bigger than 4.4mm are rejected. Out of a batch of “n” bolts 215
are accepted. How many bolts were in the batch (find n)?
X ~N(4 , 0.3)
P(3.5 < X < 4.4) = 0.05
Here
z1 = (4.4 – 4) / 0.3 = 1.3333
z2 = (3.5 – 4) / 0.3 = -1.6667
Now
P(3.5 < X < 4.4) = P(-1.67 < Z < 1.33)
= 0.90824 - .04746
= 0.86078
The number of acceptable bolts = Total Count x 0.86078
⟹
215 = n x 0.86078
⟹
n = 215 / 0.86078
⟹
249.7735 ≈ 250 bolts