Statistics - Test 3
(6.2) In a standard normal prob distribution....
MP = 0 SDP = 1
(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!
(6.2) Excel inverse find x vaule
NORM.INV
(6.2) Excel for NSD
NORM.S.DIST (assumes meand na sd)
(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 the selected consumers recognize the brand name? b. What is the probability that all of the selected consumers recognize the brand name? c. What is the probability that at least 4 of the selected consumers recognize the brand name? d. If 5 consumers are randomly selected, is 4 an unusually high number of consumers that recognize the brand name?
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.
(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 requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?
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)
(6.5) 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) 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) 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
(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.
(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)
(6.4) 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 pop medians
(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) biased estimators
median range and SD
(6.4) Unbiased estimators
sample means, sample proportions, and sample variences
(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
(6.4) The assets of the four wealthiest people are 46, 31, 23, 13. assume the same sizes are n=2 with replacment
=(A3+B1)/2 for chart at the top SMEAN = SUM OF ALL MEANS * PROBS note: PM and SM are usually equal 12122, 22121 ANNNND popMEAN = 4+31+23+13/4
(6.3) Find area of the shaded region. Graph of IQs, mean = 100, sd = 15 number on graph = 75
=NORM.DIST(75, 100, 15, TRUE)
(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. The number at the intersection of the row and column is the probability. What is the probability of the desired outcome? P(1) = 0.292 (Round to three decimal places as needed.) Therefore, the probability of 1 success with 4 trials of the experiment, given the probability 0.10 of success on a given trial, is 0.292.
(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)
(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 fir a student to pass by guessing the getting at least 6 correct answers? Why or Why not?
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.
(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) 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.
(6.3) 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)
(6.2) 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) 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 a class that runs between 50.5 and 51.75.
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)
(6.3) 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 too tall to fit through a standard doorway without bending, and what percentage of women are too tall to fit through a standard doorway without bending? b. If a statistician designs a house so that all of the doorways have heights that are sufficient for all men except the tallest 5%, what doorway height would be used?
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... closest area = 0.2005 find z-score z=-0.84 x=100+(-0.84 * 15) NOTE: 1st 2 parts are almost alwasy 0.01 and the 3rd is almost always M+(1.645*2.8)
(6.2) Find percentage of area between two z scores
1. subtract two area values from tables to find difference 2. difference * 100
(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 a particular shipment of thousands of aspirin tablets actually has a 5% rate of defects, what is the probability that this whole shipment will be accepted?
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 tablets fails to meet the required specifications, which can be written as P(X = 0) ÷P(X = 1). Now find P(X = 0) ÷ P(X =1) using the binomial probability formula. First determine the values of n, p, and q. n = 34 p= 0.05 q= 1 — 0.05= 0.95 Evaluate P(X = 0) and P(X = 1). EXCEL: [=BINOM.DIST(0, 11, 0.04, FALSE)] + [=BINOM.DIST(1, 11, 0.04, FALSE)]
(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 a particular shipment of thousands of aspirin tablets actually has a 5% rate of defects, what is the probability that this whole shipment will be accepted?
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 or 1 of the tablets fails to meet the required specifications, which can be written as P(X = 0) ÷ P(X = 1). Now find P(X = 0) ÷ P(X =1) using the binomial probability formula. First determine the values of n, p, and q. n = 34 p= 0.05 q=1 — 0.05=0.95 Evaluate P(X = 0) and P(X = 1).
(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 the min usual value (MP - 2SDP) and the max usual value (MP + SDP). n=240 p=0.75
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 =
(5.3) 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 the probability experiment represent a binomial experiment?
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 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) 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 favorably recorded
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 experiment is called a trial. 2. The trials are independent. This means the outcome of one trial will not affect the outcome of the other trials. 3. For each trial, there are two mutually exclusive (disjoint) outcomes, success or failure. 4. The probability of success is the same for each trial of the experiment. ANSWER: Yes, because the experiment satisfies all the criteria for a binomial experiment.
(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 to get at least 9 peas with green pods when 10 offspring peas are generated? Why or why not?
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 probability is one that is less than 0.05. No, because the probability of this occurring is not small.
(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 distribution table that describes the sampling distribution of the proportion of even numbers when samples of sizes n = 2 are randomly selected. Does the mean of the sample proportions equal the proportion of even numbers in the population? Do the sample proportions target the value of the population proportion? Does the sample proportion make a good estimator of the population proportion? Listed below are the nine possible samples. 4,4 4,5 4,9 5,4 5,5 5,9 9,4 9,5 9,9 gl
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
(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 unusual to get at least 10 peas with green pods when 11 offspring peas are generated? Why or why not?
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) 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 persons, not enough seats will be available. Is it unlikely for such an overbooking to occur?
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 greater than 0.05. =BINOM.DIST(22, 22, 0.89, FALSE) REMEMBER TO LOOK AT DECIMAL PLACES (4).
(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 they voted in the election. Is this result consistent with the turnout, or is this result unlikely to occur with a turnout of 74%? Why or why not? c. Based on these results, does it appear that accurate voting results can be obtained by asking voters how they acted?
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
(6.5) 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
