Statistics - Test 3

(5.3) A binomial experiment is

an experiment which satisfies these four conditions. A fixed number of trials. Each trial is independent of the others. There are only two outcomes. The probability of each outcome remains constant from trial to trial.

(5.3) Determine whether the following probability experiment represents a binomial experiment and explain the reason for your answer.
An experimental drug is administered to 100 randomly selected individuals, with the number of individuals responding favo

Recall that there are certain criteria for a binomial probability experiment. An experiment is said to be a binomial experiment provided that the following statements are true.
1. The experiment is performed a fixed number of times. Each repetition of the

(5.3) Determine whether the following probability experiment represents a binomial experiment and explain the reason for your answer. Three cards are selected from a standard 52-card deck without replacement. The number of clubs selected is recorded Does

No, because the trials of the experiment are not independent and the probability of success differs from trial to trial.

(5.3) Assume that a procedure yields a binomial distribution with n = 4 trials and a probability of success of p = 0.10. Use a binomial probability table to find the probability that the number of successes x is exactly 1.

To use a binomial probabilities table, first isolate the row with n = 4 and the corresponding value of x =1 using the labels on the right and left sides.
Next, find the column with the desired value of p = 0.10 using the labels across the top and bottom.

(5.3) Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. n= 20, x = 16,p= 0.85

P(16) = 0.182 (Round to three decimal places as needed.)

(5.3) Eleven peas are generated from parents having the green/yellow pair of genes, so there is a 0.75 probability that an individual pea will have a green pod. Find the probability that among the 11 offspring peas, at least 10 have green pods. Is it unus

The probability that at least 10 of the 11 offspring peas have green pods is 0.197 . (Round to three decimal places as needed.) Note that a small probability is one that is less than 0.05. No, because the probability of this occurring is not small.

(5.3) A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 34 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If

Like #6 BUT....Let a success be a tablet that does not meet the required specifications. To determine the correct answer, first write the desired probability as a mathematical expression. Note that this probability is equal to the probability that only 0

(5.3) A brand name has a 40% recognition rate. Assume the owner of the brand wants to verify that rate by beginning with a small sample of 5 randomly selected consumers. Complete parts (a) through (d) below.
a. What is the probability that exactly 4 of th

a. =BINOM.DIST(4, 5, 0.4, FALSE)
b. =BINOM.DIST(5, 5, 0.4, FALSE)
c. =BINOM.DIST.RANGE(5, 0.4, 4, 5)
d. No, because the probability that 4 or more of the selected consumers recognize the brand name is greater than 0.05.

(5.3) Ten peas are generated from parents having the green/yellow pair of genes, so there is a 0.75 probability that an individual pea will have a green pod. Find the probability that among the 10 offspring peas, at least 9 have green pods. Is it unusual

Same as #8 C. The probability that at least 9 of the 10 offspring peas have green pods is 0.244 . (Round to three decimal places as needed.)
Is it unusual to randomly select 10 peas and find that at least 9 of them have a green pod? Note that a small prob

(5.3) An airline has a policy of booking as many as 22 persons on an airplane that can seat only 21. (Past studies have revealed that only 89.0% of the booked passengers actually arrive for the flight.) Find the probability that if the airline books 22 pe

The probability that not enough seats will be available is 0.077 . (Round to four decimal places as needed.)
Is it unlikelyfor such an overbooking to occur? It is not unlikely for such an overbooking to occur, because the probability of the overbooking is

(5.3) A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 34 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If

Let a success be a tablet that does not meet the required specifications. To determine the correct answer, first write the desired probability as a mathematical expression. Note that this probability is equal to the probability that only 0 or 1 of the tab

(5.3) Fifteen peas are generated from parents having the green/yellow pair of genes, so there is a 0.75 probability that an individual pea will have a green pod. Find the probability that among the 15 offspring peas, no more than 1 has a green pod. Is it

(5.4) Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean (MP) and the standard deviation (SDP) Also, use the range rule of thumb to find

MP = np
SDP = sqrt(npq)
q= 1- p
SOOOO.....
MP = 240 * 0.75 = 180
q = 1- 0.75 = 0.25
SDP = sqrt(240
0.75
0.25)
MP-2SDP =
MP+2SDP =

(5.4) Several psychology students are unprepared for a surprise true/false test with 9 questions, and all of their answers are guesses. a. Find the mean and the standard deviation for the number of correct answers for such students. b. Would it be unusual

n= 9
p= 9/18 = 0.5
q= 1- 0.5 = 0.5
Then find MP, SD, MAX, and MIN
b. No, because 6 is within the range of usual values.

(5.4) A candy company claims that 20% of its plain candies are orange and a sample of 200 candies is randomly selected.

MP = 200 * 0.2
SDP = SQRT(200
0.2
0.8)
then find MIN and MAX!

(5.4) In a past election, the voter turnout was 74%. In a survey, 1066 subjects were asked if they voted in the election. a. Find the mean and standard deviation for the numbers of voters in groups of 1066. b. In the survey of 1066 people, 841 said that t

a. Mean: 1066*0.74
SD: SQRT(1066
0.74
0.26)
b. FIND MIN AND MAX. is 841 within that range?
c. No if not in range, yes if in range

(6.2) In a standard normal prob distribution....

MP = 0 SDP = 1

(6.2) The waiting times b/t a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 7 minutes. Find the prob that a randomly selected passenger has a waiting time greater that 5.25 minutes.

1. Find Length
7-0=7
2. Find area (always 1)
3. Find height
1/7 = 0.14
4. P(greater than x) = (length of shaded region) * (height of shaded region)
P=(1-5.25)*(0.14)
=(7-5.25)*(0.14)
=(1.75)*(0.14)
=0.245

(6.2) A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 49.0 and 59.0. Find the probability that a given class period runs between 50.5 and 51.75 minutes.
Find the probability of selecting

1. find general length
59-49=10
2. find length of shaded regions 51.75-50.5=1.25
3. find height 1/10=0.1
4. P= (1.25) * (0.1)

(6.2) Find the area of the sahded region. the graph depicts the SND with mean of 1 and sd of 1

(two tables) on table find value

(6.2) Assume that a randomly selected subj is given a bone density test/ Those test scores are normally distributed with a mean of 0 and a sd of 1. Find the prob that a given score is less than -1.79 and draw a sketch of the region.

To find prob:
GO TO EXCEL: NORM.DIST(-1.75, 0 , 1, TRUE)

(6.2) Assume the readings on thermometers are normally distributed with a mean of 0�C and a standard deviation of 1.00�C. Find the probability that a randomly selected thermometer reads greater than � 1.51 and draw a sketch of the region.

1. Find prob on POSITIVE table = 0.9345
In other words. You find the prob on the opposite table. if the thermometer reading = positive, look on neg table. if therm reads negative, look on positive table.

(6.2) Assume and thermometer readings are normal dist wit ha mean of 0 and a sd of 1. A thermometer is randomly selected and tested. for the case below, draw a sketch and find the prob of the reading: Between 1.25 and 2.25

1. find both areas of POSITIVE table:
1.25: 0.9878
2.25: 0.8944
2. Find diff in areas: 0.0934
DONE
NOTE: if one of the #s were -, you wld find it on -table

(6.2) Find percentage of area between two z scores

1. subtract two area values from tables to find difference
2. difference * 100

(6.2) graph of two cut-off values on a thermometer:

(6.2) Excel for NSD

NORM.S.DIST (assumes meand na sd)

(6.2) Excel inverse find x vaule

NORM.INV

(6.2) Excel for diff problems/graphs

(6.2) Cut off thermometer percents, find temp (x1 and x2)

(6.3) Find area of the shaded region. Graph of IQs, mean = 100, sd = 15 number on graph = 75

#NAME?

(6.3) Find area of the shaded region. Graph of IQs, mean = 100, sd = 15 numbers on graph = 75 and 110

(6.3) Find IQ score Mean = 100 SD=15 and area=0.4

1. find area: 1-0.4 = 0.6
2. find z-score on chart by finding number in rows and going to z-score.
3. THEN EXCEL: =NORM.INV(0.39, 100, 15) or do Mean + (x*SD)

(6.3) Find P20 of IQ scores with M of 100, SD of 15
Find nuber seperating bottom 20% with top 80% (area = 0.2)

area=low percentage 20%/100=0.2
first find area on chart... closest area = 0.2005
find z-score z=-0.84
x=100+(-0.84 * 15)

(6.3) Men's heights are normally distributed with mean 70.1 in and standard deviation of /8 in. Women's heights are normally distributed with mean 616 in and standard deviation of 15 in. The standard doorway height is 80 in. a. What percentage of men are

1. find z-score: z = (x-M)/SD
(80-70.1)/2.8=
2. find area with zscore (cumulative at the end) = 0.9999
3. 1-0.9999=0.0001
4. convert #3 to a percent=0.01
5.REPEAT FOR WOMEN
for top 5% of guys:
area=low percentage 20%/100=0.2
first find area on chart... cl

(6.3) A survey found that women's heights are normally distributed with mean 614 in and standard deviation 2.2 in. A branch of the military requires women's heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirem

a. Find z score of heights with formula (x-M)/SD
find area of each zscore
subtract
*100
b. SHORTEST 1% = area of 0.01
find on table
put into formule M+(x*SD)
TALLEST 2% = area of 0.02
find on table
put in M-(x*SD)

(6.4) Unbiased estimators

sample means, sample proportions, and sample variences

(6.4) List of number of people in a household, find median and probability of each.

c. do the sample medians target the calue of the pop mean? in general, do sample medians make good estimators of pop medians? why or why not?
ANSWER: The sample means do not target the pop median so the sample medians do not make good estimators of the po

(6.4) biased estimators

median range and SD

(6.4) The assets of the four wealthiest people are 46, 31, 23, 13. assume the same sizes are n=2 with replacment

0

(6.4) houshold populations n=2 number of ppl = 1, 3, 8 FIND VARIANCE

find =VAR.S(8,3) for all numbers. next, find var.p of pop and then add the (var*prob). Equal?

(6.4) Three randomly selected households are surveyed. The numbers of people in the households are 4, 5, and 9. Assume that samples of size n = 2 are randomly selected with replacement from the population of 4, 5, and 9. Construct a probability distributi

So if both numbers are even, prop = 1 if 1 number is even, prop = 0.5 if none are even prop = 0. soooo ex: 4,4=1

(6.5) Central Limit Theorem

(6.5) Formula for the SD of the distribution of the sample mean

(6.5) Womans heights and normally distributed with mean 63.7 and SD 1.9
a. prob of height b/t 63.3&64.3

a: =NORM.DIST(64.3, 63.7, 1.9, TRUE) - =NORM.DIST(63.3, 63.7, 1.9, TRUE)
b: 1. find new SD = SD/sqrt(n) 1.9/sqrt(10) = 0.600833
Redo NORM.DIST-NORM.DIST on EXCEL

(6.5) Elevator limit = 10 ppl. M = 180 SD 33
a. prob of weight greater than 172
b. 10 ppl greater than 172
c. does elevator have correct weight limit?

a: =1-NORM.DIST(172,180,33,TRUE)
b: 1. find new SD = SD/sqrt(n)
Redo 1-NORM.DIST on EXCEL
c: No there is a good chance their weights are over 172

(6.5) a. Boat with 60 people, prob taht their weigh exceeds limit of 145 M= 173.3 SD = 40.6
b. now 13 exceeds limit of 170?

a. find new SD = SD/sqrt(n)
Redo 1-NORM.DIST on EXCEL
b.find new SD = SD/sqrt(n)
Redo 1-NORM.DIST on EXCEL

(6.5) means of we. ight 145 lb. 60 passengers but hten change 170

a. find new SD = SD/sqrt(n)
NORM.DIST on EXCEL

(6.5) 50 passengers doors = 74 in. mean and sd of mens heights are 69 and 2.8