Analytical Methods for Engineers, Statistics and Probability

ANALYTICAL METHODS FOR ENGINEERS
STATISTICS AND PROBABILITY
Question 1:
Days lost Frequency Mid-point(Xi)
1-3 8 2
4-6 7 5
7-9 10 8
10-12 9 11
13-15 6 14
Total 40
Arithmetic Mean = {(2)(8)+(5)(7)+(8)(10)+(11)(9)+(14)(6)} /40
= (16+35+80+99+84) / 40
= 314 / 40 = 7.85
Standard Deviation = √variance
Where, Variance = [ (8)(2-7.85)^2 +(7)(5-7.85)^2+(10)(8-7.85)^2+(9)(11-7.85))+(6)(14-
7.85)] /40 = [273.78+56.8575+0.225+89.3025+226.935]/40 = 1361.52/40 = 647.1/40 =
16.1775
Therefore, Standard deviation = √(16.1775) = 4.0221
Question 2:
a. A tree diagram of the possible outcomes when a single component is selected at random.

b. i. Probability that a component was supplied by company A and was defective
P (A and D) = P (A) * P (D) = 0.60 * 0.02 = 0.012 = 1.2%
ii. Probability that the component was defective
P (A and D) or P (B and D) = P (A) * P (D) + P (B) * P (D) = (0.60*0.02) +
(0.40*0.01) = 0.012 + 0.04 = 0.052 = 5.2%
iii. Probability that the component was supplied by company A given that it was
defective
 P (A|D) = P(A and D) / P(D) = 0.012 / 0.02 = 0.6
Question 3:
X ~ N (150cm, 100cm)
a) P(X<165cm)
 P(X<165) = P(Z< ( 165−150
10 )) = P(Z < 1.5) = 0.9332
 The probability that the length of a randomly selected strip is shorter than 165 cm is
93.92%.
b) P(X>170cm)
 P(X>170) = P(Z> ( 170−150
10 )) = P(Z >2.0) = 1 – P(Z<2.0) = 1-0.9772 = 0.0228
 The probability that the length of a randomly selected strip is longer than 170cm is
2.28%
ITEM
COMPANY A
0.60
DEFECTIVE
0.02
NOT DEFECTIVE
0.98
COMPANY B
0.40
DEFECTIVE
0.01
NOT DEFECTIVE
0.99