Question

Some statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.72. Suppose that we randomly pick 25 daytime statistics students.

1. Give the distribution of X.

2. Give the distribution of

X.bar

(Round your standard deviation to three decimal places.)

3. Find the probability that an individual had between $0.68 and $1.00. (Round your answer to four decimal places.)

4. Find the probability that the average of the 25 students was between $0.68 and $1.00. (Round your answer to four decimal places.)

Answer 1

Given

Some , Statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.72

Thus X ~ Exp()

To estimate

Mean of X ; E(X) = 1 /

Given Mean of X = E(X) = 0.72

Thus    , 1 / = 0.72

                = 1 /0.72

                = 1.388889

1.

1. Give the distribution of X.

X~ Exp ( = 1.388889 )

2.

If X~ Exp ()

Then distribution of X.bar is

X.bar = =

Where ~ Gamma ( n , n* )

So if X~ Exp (=1.388889 ) , and suppose that we randomly pick n = 25 daytime statistics students.

Then distribution of X.bar ( ) is

~ Gamma ( n , n*)

where estimate of n=25 and =1.388889

3.

Find the probability that an individual had between $0.68 and $1.00. (Round your answer to four decimal places.)

i.e we need to find , P(0.68 X 1 ) = P(X 1) - P(X 0.68)

note - to find above probability we can use statistical table or any software like R

From R ( code to find P(X<=x) = " pexp(x,lamda)" )

> pexp(1,1.38889)               # P(X <= 1)
[1] 0.7506481

> pexp(0.68,1.38889)          # P(X <= 0.68)
[1] 0.6111047

Thus , P(X 1)= 0.7506481 ,   P(X 0.68) = 0.6111047

Hence required probability is

P(0.68 X 1 ) = P(X 1) - P(X 0.68)

                            = 0.7506481 - 0.6111047

P(0.68 X 1 ) = 0.1395433

Hence , probability that an individual had between $0.68 and $1.00 is 0.1395.

4. Find the probability that the average of the 25 students was between $0.68 and $1.00. (Round your answer to four decimal places.)

Here   , ~ Gamma ( n , n*)

i.e        ~ Gamma ( 25 , 34.72225)            { as n*=25*1.38889 = 34.72225 }

To find P(0.68 1 )

Thus   , P(0.68 1 ) = P( 1) - P(0.68)

note - to find above probability we can use statistical table or any software like R

From R ( code to find P(X<=x) = " pgamma(x, shape=25, rate =34.72225) " )

> pgamma(1,shape=25,rate =34.72225)                  # P( 1)
[1] 0.9642588

> pgamma(0.68,shape=25,rate =34.72225)             # P( 0.68)
[1] 0.4144748

> pgamma(1,25,34.72225)-pgamma(0.68,25,34.72225)
[1] 0.549784

Hence required probability is

P(0.68 X 1 ) = P(X 1) - P(X 0.68)

                            = 0.9642588 - 0.4144748

P(0.68 X 1 ) = 0.549784

Hence , probability that the average of the 25 students was between $0.68 and $1.00 is 0.5498.