Question

1) cars arrive at a drive-non of a bank at the rate of 4 every 10 minutes.

a - what is the probability that the next car arrives within 1 minute?
b - what is the probability that the next car arrives between 1 and 2 minutes from now?

5) For a given exam, test grades are normally distributed with mean = 70 and standard
deviation = 10.
a - When a student is selected at random, what is the probability that their grade is between
72-76?
b - what is the max grade to be in the bottom 10% of the class? (or 90% of all grades are
higher)

Answer 1

1a)here mean time of arrival =10/4 =2.5 minutes =

P( next car arrives within 1 minute )=P(X<1)=1-e^-1/2.5 =0.3297

b) probability that the next car arrives between 1 and 2 minutes from now =P(1<X<2)=(1-e^-2/2.5)-(1-e^-1/2.5)=0.2210

5)

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	70
std deviation   =σ=	10.0000

probability that their grade is between 72-76 :

probability =

P(72<X<76)

=

P(0.2<Z<0.6)=

0.7257-0.5793=

0.1464

b)

for 10th percentile critical value of z=		-1.28
therefore corresponding value=mean+z*std deviation=			57.20