Question 1 4 Marks
The stem-and-leaf display below represents the number of vitamin supplements sold by a health food store in a sample of 16 days. (1H means “high tens” – 15, 16, 17, 18, or 19.)
Stem Leaves
1H 99
2L 0023
2H 567
3L 034
3H 568
4L 1
The stem-and-leaf display below represents the number of vitamin supplements sold by a health food store in a sample of 16 days. (1H means “high tens” – 15, 16, 17, 18, or 19.)
Stem Leaves
1H 99
2L 0023
2H 567
3L 034
3H 568
4L 1
For this sample,
(a) What is the arithmetic mean of the number of vitamin supplements sold daily? 1 mark
Answer: 28
(b) What is the standard deviation? 1 mark
Answer: 7.3575
(c) Construct a box-and-whisker plot for the data in this sample. 2 marks
Answer: Five number summary; 19, 20, 26.5, 35, 41
However, Kaddstat plot gives the following (quartiles are interpolated):
(a) What is the arithmetic mean of the number of vitamin supplements sold daily? 1 mark
Answer: 28
(b) What is the standard deviation? 1 mark
Answer: 7.3575
(c) Construct a box-and-whisker plot for the data in this sample. 2 marks
Answer: Five number summary; 19, 20, 26.5, 35, 41
However, Kaddstat plot gives the following (quartiles are interpolated):
Boxplot Output
First Quartile 21.0000
Median 26.5000
Third Quartile 34.5000
Interquartile Range 13.5000
First Quartile 21.0000
Median 26.5000
Third Quartile 34.5000
Interquartile Range 13.5000
Moderate Outliers ( )
0
0
Extreme Outliers ( )
0
0
Question 2 4 Marks
Mothers Against Drunk Driving is a very visible group whose main focus is to educate the public about the harm caused by drunk drivers. A study was recently done that emphasized the problem we all face with drinking and driving. Two hundred and forty accidents that occurred on a Saturday night were analysed. Two items noted were the number of vehicles involved and whether alcohol played a role in the accident.
Mothers Against Drunk Driving is a very visible group whose main focus is to educate the public about the harm caused by drunk drivers. A study was recently done that emphasized the problem we all face with drinking and driving. Two hundred and forty accidents that occurred on a Saturday night were analysed. Two items noted were the number of vehicles involved and whether alcohol played a role in the accident.
The numbers are shown below:
Number of vehicles involved
Did alcohol play a role? 1 2 3 Totals
Yes 40 60 10 110
No 25 85 20 130
Totals 65 145 30 240
Did alcohol play a role? 1 2 3 Totals
Yes 40 60 10 110
No 25 85 20 130
Totals 65 145 30 240
An accident is chosen at random. Find the probability that, the accident
(a) involved more than one vehicle. 1 mark
Answer: (145+30)/240 = 175/240 = 72.92%
(b) involved alcohol and a single vehicle. 1 mark
Answer: 40/240 = 16.67%
(c) given that multiple vehicles were involved, alcohol was present in blood. 1 mark
Answer: (60+10)/(145+30) = 70/175 = 0.40
(d) given that alcohol was not involved, was a single vehicle accident. 1 mark
Answer: 25/130 = 0.1923
(a) involved more than one vehicle. 1 mark
Answer: (145+30)/240 = 175/240 = 72.92%
(b) involved alcohol and a single vehicle. 1 mark
Answer: 40/240 = 16.67%
(c) given that multiple vehicles were involved, alcohol was present in blood. 1 mark
Answer: (60+10)/(145+30) = 70/175 = 0.40
(d) given that alcohol was not involved, was a single vehicle accident. 1 mark
Answer: 25/130 = 0.1923
Question 3 4 Marks
The service manager for a new automobile dealership reviewed dealership records of the past 20 sales of new cars to determine the number of warranty repairs he will be called on to perform in the next 90 days. Corporate reports indicate that the probability any one of their new cars needs a warranty repair in the first 90 days is 0.05. The manager assumes that calls for warranty repair are independent of one another. What is the probability that no more than 2 cars will be needing warranty repairs to be performed in the next 90 days for this batch of 20 new cars sold?2 marks
Answer: Given, p = 0.05, n = 20, X = 0, 1, 2
The required probability = P(X=0)+P(X=1)+P(X=2) = 0.3585+0.3774+0.1887 = 0.9245
The service manager for a new automobile dealership reviewed dealership records of the past 20 sales of new cars to determine the number of warranty repairs he will be called on to perform in the next 90 days. Corporate reports indicate that the probability any one of their new cars needs a warranty repair in the first 90 days is 0.05. The manager assumes that calls for warranty repair are independent of one another. What is the probability that no more than 2 cars will be needing warranty repairs to be performed in the next 90 days for this batch of 20 new cars sold?2 marks
Answer: Given, p = 0.05, n = 20, X = 0, 1, 2
The required probability = P(X=0)+P(X=1)+P(X=2) = 0.3585+0.3774+0.1887 = 0.9245
The local police department must write, on average, 5 tickets a day to keep department revenues at budgeted levels. Suppose the number of tickets written per day follows a Poisson distribution with a mean of 6.4 tickets per day. Find the probability that less than 6 tickets are written on a randomly selected day from this population. 2 marks
Answer: P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) = 0.0017 + 0.0106 + 0.0340 + 0.0726 + 0.1162 + 0.1487 = 0.3837
Question 4 4 Marks
In the game Wheel of Fortune, the numbers 1 to 100 are uniformly distributed along the circumference of the disc, each number has the same length on the circumference as any other number. What is the probability of getting a number between 61 and 85 when the wheel is turned? 1 mark
Answer: (85 – 60)/(100 – 0) = 25/100 = 0.25
A catalogue company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 2.8 minutes. What is the probability that a randomly selected caller is placed on hold fewer than 5 minutes? 1 mark
Answer:
P(X<5) = 1 – P(x>5) = 1 – exp[(-1/2.8)5] = 1 – 0.1677 = 0.8323
The manager of a surveying company believes that the average number of phone surveys completed per hour by her employees has a normal distribution. She takes a sample of 15 days output from her employees and determines the average number of surveys per hour on these days. The data is:
10.0, 11.2, 11.4, 12.5, 12.2, 12.0, 11.5, 11.7, 11.8, 10.1, 10.3, 10.5, 10.7, 12.2, 15.0
Construct a normal probability plot and infer if the data can be considered as normally distributed. 2marks
Answer: Calculations -
Data Array Rank Percentile Z value
10 10 1 0.0625 -1.53412
11.2 10.1 2 0.125 -1.15035
11.4 10.3 3 0.1875 -0.88715
12.5 10.5 4 0.25 -0.67449
12.2 10.7 5 0.3125 -0.48878
12 11.2 6 0.375 -0.31864
11.5 11.4 7 0.4375 -0.15731
11.7 11.5 8 0.5 -1.4E-16
11.8 11.7 9 0.5625 0.157311
10.1 11.8 10 0.625 0.318639
10.3 12 11 0.6875 0.488776
10.5 12.2 12 0.75 0.67449
10.7 12.2 13 0.8125 0.887147
12.2 12.5 14 0.875 1.150349
15 15 15 0.9375 1.534121
Answer: P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) = 0.0017 + 0.0106 + 0.0340 + 0.0726 + 0.1162 + 0.1487 = 0.3837
Question 4 4 Marks
In the game Wheel of Fortune, the numbers 1 to 100 are uniformly distributed along the circumference of the disc, each number has the same length on the circumference as any other number. What is the probability of getting a number between 61 and 85 when the wheel is turned? 1 mark
Answer: (85 – 60)/(100 – 0) = 25/100 = 0.25
A catalogue company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 2.8 minutes. What is the probability that a randomly selected caller is placed on hold fewer than 5 minutes? 1 mark
Answer:
P(X<5) = 1 – P(x>5) = 1 – exp[(-1/2.8)5] = 1 – 0.1677 = 0.8323
The manager of a surveying company believes that the average number of phone surveys completed per hour by her employees has a normal distribution. She takes a sample of 15 days output from her employees and determines the average number of surveys per hour on these days. The data is:
10.0, 11.2, 11.4, 12.5, 12.2, 12.0, 11.5, 11.7, 11.8, 10.1, 10.3, 10.5, 10.7, 12.2, 15.0
Construct a normal probability plot and infer if the data can be considered as normally distributed. 2marks
Answer: Calculations -
Data Array Rank Percentile Z value
10 10 1 0.0625 -1.53412
11.2 10.1 2 0.125 -1.15035
11.4 10.3 3 0.1875 -0.88715
12.5 10.5 4 0.25 -0.67449
12.2 10.7 5 0.3125 -0.48878
12 11.2 6 0.375 -0.31864
11.5 11.4 7 0.4375 -0.15731
11.7 11.5 8 0.5 -1.4E-16
11.8 11.7 9 0.5625 0.157311
10.1 11.8 10 0.625 0.318639
10.3 12 11 0.6875 0.488776
10.5 12.2 12 0.75 0.67449
10.7 12.2 13 0.8125 0.887147
12.2 12.5 14 0.875 1.150349
15 15 15 0.9375 1.534121
The plot below shows that the data do not appear to follow a normal distribution.
Question 5 4 Marks
The owner of a fish market has an assistant who has determined that the weights of Salmon are normally distributed with mean of 3.2 kg and standard deviation of 0.8 kg. If a sample of 40 fish yields a mean of 3.4 kg, what is the probability of obtaining a sample mean this large or larger? 1mark
Answer: Z_1=(3.4-3.2)/(0.8/v40)=1.58
P(X>3.4) = 1 – P(X<3.4) = 1 – P(Z<Z1) = 1 – 0.943 = 0.057
A wealthy real estate investor wants to decide whether it is a good investment to build a high-end shopping complex in a suburb of Brisbane. His main concern is the total market value of the 3,605 houses in the suburb. He commissioned a statistical consulting group to take a sample of 200 houses and obtained a sample average market price of $325,000 and a sample standard deviation of $38,700. If he wants a 95% confidence on estimating the true population average market price of the houses in the suburb to be within $10,000, how large a sample will he need? (Use finite population correction factor.) 1.5marks
Answer: n = 200, X ¯=$325,000, S=$38,700
With finite population correction factor (and since sample size is greater than 30, we can use z instead of t)
$10,000 = 1.96S/vn v((N-n)/(N-1)) = 1.9638,700/vn v((3605-n)/(3605-1)); We have to solve for n.
Squaring both sides and solving for n, we get,n =56.65 ˜ 57 (must always be rounded up)
How many Kleenex pieces the Kimberly Clark Corporation package of tissues contain? Researchers determined that 60 tissues is the average number of tissues used during a cold. Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: X ¯=52, S = 22. Suppose the alternative hypothesis we want to test isH_1: µ<60. With a = 0.05, should we reject the null hypothesis? 1.5marks
Answer: The hypotheses are:-H_0: µ=60 and H_1: µ<60. A sample size of 100 means we can use normal distribution. This is a one-tailed test and the critical value is -1.645. The estimated z value is (52 – 60)/(22/v100)= -3.64. Since the estimated value is more extreme than the critical value, we reject the null hypothesis
The owner of a fish market has an assistant who has determined that the weights of Salmon are normally distributed with mean of 3.2 kg and standard deviation of 0.8 kg. If a sample of 40 fish yields a mean of 3.4 kg, what is the probability of obtaining a sample mean this large or larger? 1mark
Answer: Z_1=(3.4-3.2)/(0.8/v40)=1.58
P(X>3.4) = 1 – P(X<3.4) = 1 – P(Z<Z1) = 1 – 0.943 = 0.057
A wealthy real estate investor wants to decide whether it is a good investment to build a high-end shopping complex in a suburb of Brisbane. His main concern is the total market value of the 3,605 houses in the suburb. He commissioned a statistical consulting group to take a sample of 200 houses and obtained a sample average market price of $325,000 and a sample standard deviation of $38,700. If he wants a 95% confidence on estimating the true population average market price of the houses in the suburb to be within $10,000, how large a sample will he need? (Use finite population correction factor.) 1.5marks
Answer: n = 200, X ¯=$325,000, S=$38,700
With finite population correction factor (and since sample size is greater than 30, we can use z instead of t)
$10,000 = 1.96S/vn v((N-n)/(N-1)) = 1.9638,700/vn v((3605-n)/(3605-1)); We have to solve for n.
Squaring both sides and solving for n, we get,n =56.65 ˜ 57 (must always be rounded up)
How many Kleenex pieces the Kimberly Clark Corporation package of tissues contain? Researchers determined that 60 tissues is the average number of tissues used during a cold. Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: X ¯=52, S = 22. Suppose the alternative hypothesis we want to test isH_1: µ<60. With a = 0.05, should we reject the null hypothesis? 1.5marks
Answer: The hypotheses are:-H_0: µ=60 and H_1: µ<60. A sample size of 100 means we can use normal distribution. This is a one-tailed test and the critical value is -1.645. The estimated z value is (52 – 60)/(22/v100)= -3.64. Since the estimated value is more extreme than the critical value, we reject the null hypothesis
No comments:
Post a Comment