Question

Some statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students.

Give the distribution of

X.

(Round your standard deviation to three decimal places.)

X

~

,

Find the probability that an individual had between $0.68 and $0.96. (Round your answer to four decimal places.)

Find the probability that the average of the 25 students was between $0.68 and $0.96. (Round your answer to four decimal places.)

Answer 1

1)probability that an individual had between $0.68 and $0.96

P(0.68<X<0.96)=(1-exp(-0.96/0.88)-(1-exp(-0.68/0.88))=

0.126

std error=σx̅=σ/√n=0.88/√25 =

0.176

distribution of Xbar~ N(0.88, 0.176)

2)

probability that the average of the 25 students was between $0.68 and $0.96.:

probability =P(0.68<X<0.96)=P((0.68-0.88)/0.176)<Z<(0.96-0.88)/0.176)=P(-1.14<Z<0.45)=0.6753-0.1279=0.5474

(please try 0.5465 if this comes wrong and reply)