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Basic Statistics Questions and Answers
Statistics:Statistic deals with the study of collection, classification and analyzing of data over a certain range of frequency distribution. In other words Statistics involves with collection of data and preparing frequency distribution table over given set of data.
a) 50 b) 60 c) 55 d) 65
2. Two consecutive class marks of a distribution are 36 and 41. Then the class size is :
a) 5 b) 4 c) 3 d) 2
a) 2m + l b) 2m  l c) m  l d) m  2l
5. The following is the distribution of marks of Arun class VI in all is subject. Draw a histogram for the above data.
6. the mean of the prime number between 40 and 50 is:
a) 41 b) 45 c) 47 d) 50
a) Median b) Mode c) Data d) Histogram
8. Find the mean of the factors of 36.
a) 11 b) 20 c) 12.5 d) 11.25
9. Find the median of first ten prime numbers
a) 12 b) 32 c) 20 d) 16
10. Find the mean of the number 2,3,5,7,11,13,17,19,23 and 29
a) 12 b) 12.5 c) 12.9 d) 13
Answer:
1. 55
Basic Statistic Question and Answer
1. Class mark of class interval 5060 is :
a) 50 b) 60 c) 55 d) 65
2. Two consecutive class marks of a distribution are 36 and 41. Then the class size is :
a) 5 b) 4 c) 3 d) 2
3. Let m be the mid value and l be the upper limit of a class in a frequency distribution. The lower limit of the class is :
a) 2m + l b) 2m  l c) m  l d) m  2l
4. The marks obtained by 40 student of class X in an examination are given below.
18,8,12,6,8,16,12,5,23,2,16,23,2,10,20,12,9,7,6,5,3,5,13,21,13,15,2024,1,7,21,16,13,18,17,16.
Represent the data in the form of a frequency distribution table of the same class size.
5. The following is the distribution of marks of Arun class VI in all is subject. Draw a histogram for the above data.
Subject  English  Maths  Science  Geography  Economics  History 
Marks  40  35  45  41  48  30 
6. the mean of the prime number between 40 and 50 is:
a) 41 b) 45 c) 47 d) 50
7. The information collected with definite reason are called:
a) Median b) Mode c) Data d) Histogram
8. Find the mean of the factors of 36.
a) 11 b) 20 c) 12.5 d) 11.25
9. Find the median of first ten prime numbers
a) 12 b) 32 c) 20 d) 16
10. Find the mean of the number 2,3,5,7,11,13,17,19,23 and 29
a) 12 b) 12.5 c) 12.9 d) 13
Answer:
1. 55
2. 5
3. 2ml
4. The data in frequency distribution form.
5. The histogram for the data is given below.
6. 45
Marks  Frequency 
05  5 
510  11 
1015  7 
1520  9 
2025  8 
Total  40 
5. The histogram for the data is given below.
6. 45
7. Data
8. The factors of 36 are 2,18,3,12,4,9,6,36
Mean of the factor = $\frac{2,3,4,6,9,12,18,36}{8}$
= $\frac{90}{8}$ = 11.25
9. First 10 prime numbers are 2,3,5,711,13,17,19,23 and 29.
number of terms are 10.
mean of term is $\frac{(11+13)}{2}$ = 12.
10. The mean of the number 2,3,5,7,11,13,17,19,23 and 29
= $\frac{129}{10}$ = 12.9
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From Yahoo Answers
Question:Edexcel Statistics 1 January 2008 Exam:
Question 2 B )
You don't really need to see the question, but basically I've ordered the data and get:
125 160 169 171 175 186 210 243 250 258 390 420
12 numbers therefore mediam is n+1 divided by 2 and i got 6.5 so basically i took 186+210 divided by 2 getting 198. That's fine!
Next I have to find the quartiles:
For Q1 I said n divided by 4 is 3 and took the thrid term 169
and for Q3 n divided by 4 times 3 is the 9th term and is 250
Both of these were wrong?
For some reason the got 170 for Q1 and 254 for Q3
I can see where they have got the answer from (by averaging) but I don't understand. The rules in the book tell us that if we have "n" terms we do "n" divided by 4 and round up..e.g for 11 terms I would get 11 divided by 4 giving 2.75 and take the 3rd term. However I am unsure as to what they have done in this exam?
My method had worked for the previous pastpapers but for this one it hasn't.
Your help is greatly appreciated. Thanks, but on the mark scheme they only accepted the other method, i guess they may have accepted mine but didn't necessarily write it on the actual mark scheme.
Answers:basically your own previous work is right. The second method is usually used for university students
Answers:basically your own previous work is right. The second method is usually used for university students
Question:Hi there, my mom who is taking this class is having difficulty solving a few problems. She is currently going to a 6 week class that doesn't offer any math tutoring. Since I took this course years ago I don't remember anything about it so I would appreciate some help. Thanks everyone!
14. Foofy has a normal distribution of scores ranging from 2 to 9.
This actually confused me since I remember about range limits and range, but nothing about "ranging from".
(a) She computed the variance to be 0.6. What should you conclude from this answer, and why?
(b) She recomputes the standard deviation to be 18. What should you conclude, and why?
(c) She recomputes the variance to be 1.36. What should you conclude, and why?
There is a lot of problems that go what this mean and so forth. Can anyone give me some pointers in understanding this better.
If someone can offer a site that has statistics guides and outlines, that would be most helpful. I've tried looking but have found only websites for everything but statistics.
Answers:(a) She computed the variance to be 0.6 Is this negative 0.6? Variance cannot be negative, The least possible value is 0 which happens when all of the observations are equal. b) Standard deviation seems to be too large for values ranging from 2 to 9. This also implies variance is 18 x 18 = 324 (unlikely) c) 1.36 is possible. it means a small variance and the observations are close to the mean of the observations. Also, the standard deviation is square root of 1.36 = 1.167
Answers:(a) She computed the variance to be 0.6 Is this negative 0.6? Variance cannot be negative, The least possible value is 0 which happens when all of the observations are equal. b) Standard deviation seems to be too large for values ranging from 2 to 9. This also implies variance is 18 x 18 = 324 (unlikely) c) 1.36 is possible. it means a small variance and the observations are close to the mean of the observations. Also, the standard deviation is square root of 1.36 = 1.167
Question:Hey Everyone! I'm working on a project for my statistics class and I've got most all of my project done (this is literally the last bullet of about 50 things on the assignment) and I'm having some trouble with figuring out exactly how to complete this! So if anyone could help me understand it, or at least give me some direction, I'd really appreciate it!
Basically, we had to create a binomial event by questioning 50 fellow students a y/n question. I considered success being an answer of "yes" 50 times. The results were 34 "yes" and 16 "no". Now, here is where I get confused: "Generate three binomial probability questions related to your event that must be solved using Excel. Solve these questions using Excel and present the algorithm you used along with the results."
Again, I'm not asking for anyone to do my homework for me, just a little guidance and tutoring would help immensely!
Answers:So, "success" in this case, is a "yes" answer on any particular trial. "Failure" is a "no" answer on any particular trial. With this in mind, p (the probability of success on any particular trial) is 34/50, or .68; meaning that q (probability of failure on any particular trial) is 16/50, or .32. Now, I believe you're being asked to generate a questions saying something like "What is the probability of greater than 30 students answering yes?" In this case, you can use excel to calculate the probability of each number of successes within the range (3150, in this case), and have excel sum them all. To calculate a particular number of successes (r) out of a certain number of trials (n) with probability of success on any particular trial (p), use the formula: P(r successes) = p^r * q^(nr) * nCr You'll have Excel run one of these for each r, so I'll show one for example, 31 successes: P(31 successes) = .68^31 * .32^19 * 50C31 Good luck.
Answers:So, "success" in this case, is a "yes" answer on any particular trial. "Failure" is a "no" answer on any particular trial. With this in mind, p (the probability of success on any particular trial) is 34/50, or .68; meaning that q (probability of failure on any particular trial) is 16/50, or .32. Now, I believe you're being asked to generate a questions saying something like "What is the probability of greater than 30 students answering yes?" In this case, you can use excel to calculate the probability of each number of successes within the range (3150, in this case), and have excel sum them all. To calculate a particular number of successes (r) out of a certain number of trials (n) with probability of success on any particular trial (p), use the formula: P(r successes) = p^r * q^(nr) * nCr You'll have Excel run one of these for each r, so I'll show one for example, 31 successes: P(31 successes) = .68^31 * .32^19 * 50C31 Good luck.
Question:The Environmental protection Agency sets limits on the maximum allowable concentration of certain chemicals in drinking water from wells. For the substance PCB, the limit has been set at 5 ppm. A random sample of 50 water specimens from the same well results in 3 specimens with a PCB concentration exceeding 5ppm.
Give a 85% confidence interval for the proportion of wells with PCB levels exceeding 5ppm.
10 points to the best answer. (Show how you got to this)
Answers:ANSWER: 85% Resulting Confidence Interval for 'true mean' for water specimens exceeding 5ppm = [0.01, 0.11] Why??? POPULATION PROPORTION, CONFIDENCE INTERVAL, NORMAL DISTRIBUTION n: Number of samples = 50 k: Number of Successes = 3 p: Sample Proportion [3/50] = 0.06 significant digits: 2 Confidence Level = 85 "Lookup" Table 'zcritical value' = 1.440 This 'zcritical value' is a Lookup from the Table of Standard Normal Distribution. The Table is organized as a cummulative 'area' from the LEFT corresponding to the STANDARDIZED VARIABLE z. The Standard Normal Distribution is symmetric (called a 'Bell Curve') which means its an interpretive procedure to LookUp the 'area' from the Table. For the Confidence Level (or Level of Confidence) = 85, there is a LEFT 'area' OUTSIDE. And due to symmetry there is a RIGHT 'area' OUTSIDE. Using a Lookup from the Table involves adding and subtracting an 'area' which is equal to the Confidence Level. For STANDARDIZED VARIABLE z = 1.44, this corresponds to the LEFT 'area' half of the Confidence Level area = 0.5 * (1  85/100) = 0.08 by a Lookup in the Table for Standard Normal Distribution. Or alternative; use Excel function: NORMSINV(probability) Returns the inverse of the standard normal cumulative distribution. The distribution has a mean of zero, a standard deviation of one and is symmetrical. 85% Resulting Confidence Interval for 'true mean': p +/ ('z critical value') * SQRT[p * (1  p)/n] = 0.06 +/ 1.44 * SQRT[0.06 * (1  0.06)/50] = [0.01, 0.11]
Answers:ANSWER: 85% Resulting Confidence Interval for 'true mean' for water specimens exceeding 5ppm = [0.01, 0.11] Why??? POPULATION PROPORTION, CONFIDENCE INTERVAL, NORMAL DISTRIBUTION n: Number of samples = 50 k: Number of Successes = 3 p: Sample Proportion [3/50] = 0.06 significant digits: 2 Confidence Level = 85 "Lookup" Table 'zcritical value' = 1.440 This 'zcritical value' is a Lookup from the Table of Standard Normal Distribution. The Table is organized as a cummulative 'area' from the LEFT corresponding to the STANDARDIZED VARIABLE z. The Standard Normal Distribution is symmetric (called a 'Bell Curve') which means its an interpretive procedure to LookUp the 'area' from the Table. For the Confidence Level (or Level of Confidence) = 85, there is a LEFT 'area' OUTSIDE. And due to symmetry there is a RIGHT 'area' OUTSIDE. Using a Lookup from the Table involves adding and subtracting an 'area' which is equal to the Confidence Level. For STANDARDIZED VARIABLE z = 1.44, this corresponds to the LEFT 'area' half of the Confidence Level area = 0.5 * (1  85/100) = 0.08 by a Lookup in the Table for Standard Normal Distribution. Or alternative; use Excel function: NORMSINV(probability) Returns the inverse of the standard normal cumulative distribution. The distribution has a mean of zero, a standard deviation of one and is symmetrical. 85% Resulting Confidence Interval for 'true mean': p +/ ('z critical value') * SQRT[p * (1  p)/n] = 0.06 +/ 1.44 * SQRT[0.06 * (1  0.06)/50] = [0.01, 0.11]
From Youtube
The Basics: Descriptive and Inferential Statistics :statisticslectures.com  where you can find free lectures, videos, and exercises, as well as get your questions answered on our forums!
Basic Math: Lesson 6  Answers to Quiz Questions :Stepbystep answers to the quiz that is at the end of the Lesson Notes.