1. suppose you have 64 independent random variables all with an exponential distribution with mean 5 and variance 25 . let x denote the sample mean of the 64 random variables. a) what is the distribution of x ? b) what is the probability p(x > 6) c) what is the 90th percentile of x ? d) what is the 10th percentile of x ?

0.01506, 6.06667,3.9334

Step-by-step explanation:

Given that you have 64 independent random variables all with an exponential distribution with mean 5 and variance 25 .

Let X denote the sample mean of the 64 random variables.

The sample mean i.e. X will have a normal distribution irrespective of the original parent distribution.  

This is because of central limit theorem.  The central limit theorem says that for large sample sizes randomly drawn, the sample mean would follow a normal distribution irrespective of the distribution of the population

X is N(5,

a) X is N(5, 0.8333)

b)

c) 90th percentile of X =

d) 10th percentile =

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

your answer is -9

step-by-step explanation: