Question

Question 5

a) (1) X~Normal(mean=4, standard deviation=3), (2) Y~Normal(mean=6, standard deviation = 4), and (3) X and Y are independent, then, P(X+Y>13) equals (in 4 decimal places)

Answers options: a) 0.7257, b) 0.3341, c) 0.2743, d) 0.6759, e) none of these

b) Let X~Gamma(4, 1.2). Which of the following is possible R code for computing the probability that X < 2.6?

Answers options: a) dgam(2.6, 4, 1.2), b) pgamma(4, 1.2, 2.6), c) dgamma(2.6, 4, 1.2), d) pgamma(2.6, 4, 1.2), e) None of these

c) If X~Exponential(lambda=2.8), which of the following code computes P(X>2) correctly?

answers options: a) 1-dexp (2, rate=2.8), b) pexp(2, rate=2.8), c) 1-pexp(2, rate=2.8), d) dexp(2, rate=2.8)

d) If X has an Exponential distribution with mean 2.5, which of the following code computes P(X<3) correctly?  

Answers options: a) pexp(3, rate=2.5), b) dexp(3, rate=2.5), c) pexp(3, rate=0.4), d) dexp(3, rate=0.4), e) None of these

e) If X1, X2, ..., X100 are independent and identically distributed as Uniform (0,1), the probability that the average of these 100 random variables is less than 0.3 equals   , approximately? ( ). Answers options: a) 0.4586, b) 0.5414, c) 0.6406, d) 0.3594, e) None of these

Answer 1

Solution-:

(a) Here, we used additive property of normal distribution

If and , X and Y are independent then

Where, and

By using R-Software:

> #(a) We find P[Z=X+Y>13]
> p=1-pnorm(13,10,5);p   
[1] 0.2742531
> round(p,4)   
[1] 0.2743

Option (c) 0.2743 is correct

(b) Let

By using R-software :

>#(b) We find P[X<2.6]

> pgamma(2.6, 4, 1.2)
[1] 0.3796321
Option (d) pgamma(2.6, 4, 1.2)   is correct

(c) Let

By using R-software :

> #(c)We find P[X>2]
> p1=1-pexp(2,2.8);p1
[1] 0.003697864

Option (c) 1-pexp(2, rate=2.8) is correct

(d) Let

where,

> #(d)We find P[X<3]
> p2=pexp(3,0.4);p2
[1] 0.6988058
Option (c) pexp(2, rate=0.4) is correct