Assignment
STA -301
Question 1 A: If Z is a standard normal random variable with mean 0 and variance 1, then find the lower quartile?
Solution:
Means, µ =0
Variance 62 = 1
Q1 = µ - 0.6745 б
Q1 = 0 - .6745 (1)
Q1 = -.6745
Question 1 C: What is probability that a poker hand of 5cards contain exactly 1 ace?
Solution:
Probability of exactly one ace in poker hand of 5cards
=
= 4 X 194580
2598960
=778320 .
2598960
=0.2994
Question 1 D: A certain type of storage battery lasts on the average 3.0 years, with a standard deviation of 0.5 year. Assuming that the battery lives are normally distributed, find the probability that a given battery will last less than 2.3 years.
Solution:
Here
Mean µ =3.0
S.D =0.5
P(x 2.3) = P Z 2.3-3.0
0.5
Standardizing
P(x 2.3) = P Z -1.4
=P(-0 Z 0)-P(0 Z -1.4)
=0.5-P(0 Z 1.4)
=0.5-0.4192
=0.0808 Answer
Question 1 E: The incident of occupational disease is such that the workers have 20 percent chance of suffering from it. What is the probability that out of six more than 4 workers will come in contact of the disease?
Solution:
Here
P=20 .
100
=0.2
q= 1-P
=1-0.2
=0.8
n = 6
P(more than 4) = P(x 4) = P(x=5)+P(x=6)
By Binomial Formula
P(x = x)= pxqn-x
P(x 4) = (0.2)5X(0.8)6-5+ (0.2)6X(0.8)6-6
P(x 4) =6.5.4.3.2.1 (0.2)5X(0.8)6-5+6.5.4.3.2.1(0.2)6X(0.8)6-6
5.4.3.2.1 6.5.4.3.2.1
P(x 4) = 6X(0.2)5X(0.8)1+1X(0.2)6X1
P(x 4) =6X3.2X0.8+1X6.4X1
P(x 4) =15.36+6.4
P(x 4) =21.76Answer
Question 2A: The Probability distribution of X is given below:
Compute,
i) P(x=2)
ii) E(X)
| X | 0 1 2 3 4 | ||
| P(X) |
|
1: P(x=2)
P(x=2)
=35/100
=7/20
| X | P(X) | XP(X) | |||
| 0 | 1/10 | 0 | |||
| 1 | ½ | ½ | |||
| 2 | 35/100 | 70/100 | |||
| 3 | 4/100 | 12/100 | |||
| 4 | 1/100 | 4/100 |
2: E(X) = XP(X) = 0 + 1 + 70 + 12 + 4
2 100 100 100
= 50+70+12+4
100
= 136
100
= 1.36
Question 2B: The p. d. f. of a random variable is given
f(x) = (2x-1) 1<x < 2
0 Elsewhere
Calculate P(X<1.5).
Solution:
F(x) = 2x-1 1<x < 2
P(x 1.5) = 1 1.5 (2x-1) dx
P(x 1.5) =1 1.5 x dx - 1 1.5 1 dx
P(x 1.5) = 1(2x2)1.5 – 1(x)1.5
P(x 1.5) = ((1.5)2 – (1)2) – (1.5 -1)
P(x 1.5) = 2.25 – 1 – 0.5
P(x 1.5) = 1.25 – 0.5
P(x 1.5) = 0.75
Assalam u Alaikum.
Any body send me the solution of STA301 assignment 3.Today is last date....
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