Find the probability that her cats will wake her up no more than 5 times next week.
0.5000
0.9329
0.0378
0.0671
D: 0.0671
People visiting video rental stores often rent more than one DVD at a time.
The probability distribution for DVD rentals per customer at Video To Go is given below. There is 5 video limit per customer at this store, so nobody ever rents more than 5 DVDs.
x
0
1
2
3
4
5
P(X=x)
0.03
0.50
0.24
?
0.07
0.04
Describe the random variable X in words.
Find the probability that a customer rents three DVDs.
Find the probability that a customer rents at least 4 DVDs.
Find the probability that a customer rents at most 2 DVDs.
Another shop, Entertainment Headquarters, rents DVDs and videogames. The probability distribution for DVD rentals per customer at this shop is given below. They also have a 5 DVD limit per customer.
x
0
1
2
3
4
5
P(X=x)
0.35
0.25
0.20
0.10
0.05
0.05
At which store is the expected number of DVDs rented per customer higher?
If Video to Go estimates that they will have 300 customers next week, how many DVDs do they expect to rent next week? Answer in sentence form.
If Video to Go expects 300 customers next week and Entertainment HQ projects that they will have 420 customers, for which store is the expected number of DVD rentals for next week higher? Explain.
Which of the two video stores experiences more variation in the number of DVD rentals per customer? How do you know that?
Partial Answer:
A: X = the number of DVDs a Video to Go customer rents
B: 0.12
C: 0.11
D: 0.77
A game involves selecting a card from a deck of cards and tossing a coin.
The deck has 52 cards and 12 cards are "face cards" (Jack, Queen, or King)The coin is a fair coin and is equally likely to land on Heads or Tails
If the card is a face card and the coin lands on Heads, you win $6
If the card is a face card and the coin lands on Tails, you win $2
If the card is not a face card, you lose $2, no matter what the coin shows.
Find the expected value for this game (expected net gain or loss).
Explain what your calculations indicate about your long-term average profits and losses on this game.
Should you play this game to win money?
The variable of interest is X = net gain or loss, in dollars
The face cards J, Q, K (Jack, Queen, King).
There are(3)(4) = 12 face cards and 52 – 12 = 40 cards that are not face cards.
We first need to construct the probability distribution for X.
We use the card and coin events to determine the probability for each outcome, but we use the monetary value of X to determine the expected value.
Expected value = –$0.62, rounded to the nearest cent
If you play this game repeatedly, over a long number of games, you would expect to lost 62 cents per game, on average.
You should not play this game to win money because the expected value indicates an expected average loss.
You buy a lottery ticket to a lottery that costs $10 per ticket. There are only 100 tickets available be sold in this lottery. In this lottery there is one $500 prize, 2 $100 prizes and 4 $25 prizes.
Find your expected gain or loss.
Start by writing the probability distribution.
X is net gain or loss = prize (if any) less $10 cost of ticket
X = $ net gain or loss
P(X)
$500–$10=$490
1/100
$100–$10=$90
2/100
$25–$10=$15
4/100
$0–$10=$–10
93/100)
Expected Value = (490)(1/100) + (90)(2/100) + (15)(4/100) + (–10) (93/100) = –$2.There is an expected loss of $2 per ticket, on average.
A student takes a 10 question true-false quiz, but did not study and randomly guesses each answer.
Find the probability that the student passes the quiz with a grade of
at least 70% of the questions correct.
X = number of questions answered correctly
X~B(10, 0.5)
We are interested in AT LEAST 70% of 10 questions correct. 70% of 10 is 7. We want to find the probability that X is greater than or equal to 7.
The event "at least 7" is the complement of "less than or equal to 6".
Using your calculator's distribution menu:
1 – binomcdf(10, .5, 6) gives 0.171875
The probability of getting at least 70% of the 10 questions correct when randomly guessing is approximately 0.172
A student takes a 32 question multiple choice exam, but did not study and randomly guesses each answer. Each question has 3 possible choices for the answer. Find the probability that the student guesses
more than 75% of the questions correctly.
X = number of questions answered correctly
X~B(32, 1/3)
We are interested in MORE THAN 75% of 32 questions correct. 75% of 32 is 24. We want to find P(x>24).
The event "more than 24" is the complement of "less than or equal to 24".
Using your calculator's distribution menu:
1 - binomcdf(32, 1/3, 24)
P(x>24) = 0.00000026761
The probability of getting more than 75% of the 32 questions correct when randomly guessing is very small and practically zero.
Suppose that you are perfoming the probability experiment of rolling one fair six-sided die. Let F be the event of rolling a "4" or a "5". You are interested in how many times you need to roll the die in order to obtain the first “4 or 5” as the outcome.
p = probability of success (event F occurs)
q = probability of failure (event F does not occur)
Write the description of the random variable X. What are the values that X can take on? Find the values of p and q.
Find the probability that the first occurrence of event F (rolling a “4” or “5”) is on the second trial.
How many trials would you expect until you roll a “4” or “5”?
A: X can take on the values 1, 2, 3, .... p = 2/6, q = 4/6
B: 0.2222
C: 3
**Exercises 38 - 43 contributed by Roberta Bloom
Questions & Answers
can someone help me with some logarithmic and exponential equations.
In this morden time nanotechnology used in many field .
1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc
2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc
3- Atomobile -MEMS, Coating on car etc.
and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change .
maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Source:
OpenStax, Collaborative statistics (custom lecture version modified by t. short). OpenStax CNX. Jul 15, 2013 Download for free at http://cnx.org/content/col11543/1.1
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