The lengths of time bank customers must wait for a teller are normally distribut
ID: 3206435 • Letter: T
Question
The lengths of time bank customers must wait for a teller are normally distributed, with a mean of 3 minutes and a standard deviation of 2 minute.
What proportion of bank customers waits between 3 and 4.5 minutes
What percentage wait more than 4 minutes?
What proportion waits between 2 and 3.5 minutes?
What percentage wait less than 1 minute?
2. To estimate the medical charges for an appendectomy Blue Star Insurance has data from a random
sample of 75 patients. The sample mean cost is $410, with a sample standard deviation of $90.
Construct a 95% confidence interval for the population mean cost.
We are 95% confident that the populations mean costs of patients’ medical charges for an appendectomy through Blue Star Insurance range is between _____________ and ________________
3.
Democrat
Republican
Independent
DC
10
15
65
Los Angeles
35
25
75
North Carolina
60
20
70
A group of voters from 3 states were selected and the results are above.
What is the probability of selecting a random voter from Los Angeles, given that the voter was a republican?
a).357
b).417
c).267
d).200
4..
Democrat
Republican
Independent
Maryland
10
15
65
Los Angeles
35
25
75
North Carolina
60
20
70
A group of voters from 3 states were selected and the results are above.
What is the probability of selecting a random voter that is from North Carolina or a Democrat?
a).093
b).520
c).613
d).357
5. In 20 pulls from a deck of playing cards, 20 cards were red. What is the probability that on the 21st pull the card will be red?
a).5000
b).2500
c).000000477
d).03125
Democrat
Republican
Independent
DC
10
15
65
Los Angeles
35
25
75
North Carolina
60
20
70
Explanation / Answer
1) here from normal distribution Z=(X-mean)/std deviation
a)hence proportion of bank customers waits between 3 and 4.5 minutes=P(3<X<4.5)=P((3-3)/2<Z<(4.5-3)/2)
=P(0<Z<0.75) =0.7734-0.5=0.2734
b) P(X>4)=1-P(X<4) =1-P(Z<(4-3)/2)=1-P(Z<0.5)=1-0.6915=0.3085
hence 30.85% wait for more then 4 minutes
c)P(2<X<3.5)=P(-0.5<Z<0.25)=0.5987-0.3085 =0.2902
d)P(X<1)=P(Z<-1)=0.1587
hence 15.87% wait less then 1 minute
2) std error=(std deviation/(n)1/2 =90/(75)1/2 =10.3923
for 95% CI, z=1.96
hence confidence interval =mean +/- z*std deviation =389.63 to 430.37
3)probability =25/(15+25+20)=25/60=0.417 option B
4)total candidates =375
hence probability =(10+35+60+20+70)/375=0.52 option B
5)as it is an independent event hence probabilty =26/52=0.5
option a
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.