course Mth152 ~Rΰĕ|P|~assignment #014
......!!!!!!!!...................................
07:32:34 `q001. Note that there are 8 questions in this assignment. {}{} What is the average, or mean value, of the numbers 5, 7, 9, 9, 10, 12, 13, and 15? On the average how 'far' is each number from this mean value?
......!!!!!!!!...................................
RESPONSE --> you first add 5+7+9+9+10+12+13+15=80, the next step is you will divide by 8 the mean is 80/8=10 The avg. differences are 5+3+1+1+0+2 +3+5/8=20 20/8=2.5 The way you come up with this is you will take each #(5,7,9,12,13,and 15) and the mean of 10, which gives the above list 5,3,1,1,2,3,5 confidence assessment: 3
.................................................
......!!!!!!!!...................................
07:34:44 `q002 What is the middle number among the numbers 13, 12, 5, 7, 9, 15, 9, 10, 8?
......!!!!!!!!...................................
RESPONSE --> 9 is the middle # confidence assessment: 3
.................................................
......!!!!!!!!...................................
07:36:00 `q003. On a list of 9 numbers, which number will be the one in the middle? Note that the middle number is called the 'median'.
......!!!!!!!!...................................
RESPONSE --> ? confidence assessment: 0
.................................................
......!!!!!!!!...................................
07:37:38 `q004. What is the median (the middle number) among the numbers 5, 7, 9, 9, 10, 12, 13, and 15?
......!!!!!!!!...................................
RESPONSE --> 9.5 b/c there are 2 middle #s 9 and 10, and you cant have both of them confidence assessment: 3
.................................................
......!!!!!!!!...................................
07:39:32 `q005. We saw that for the numbers 5, 7, 9, 9, 10, 12, 13, and 15, on the average each number is 2.5 units from the average. Are the numbers in the list 48, 48, 49, 50, 51, 53, 54, 55 closer or further that this, on the average, from their mean?
......!!!!!!!!...................................
RESPONSE --> 48+48+49+50+51+53+54+55/8=408 408/8=51 closer confidence assessment: 2
.................................................
......!!!!!!!!...................................
07:43:28 `q006. On a 1-10 rating of a movie, one group gave the ratings 1, 8, 8, 9, 9, 10 while another gave the ratings 7, 7, 8, 8, 9, 10. Find the mean (average) and the median (middle value) of each group's ratings. Which group would you say liked the movie better?
......!!!!!!!!...................................
RESPONSE --> Group 1 : 1+8+8+9+9+10/6=45 45/6=7.5 8.5 is the median Group 2: 7+7+8+8+9+10/6=49 49/6=8.16 8 is the median Group 1 liked it more. confidence assessment: 3
.................................................
......!!!!!!!!...................................
07:45:28 `q007. Suppose that in a certain office that ten employees make $700 per pay period, while five make $800 per pay period and the other two make $1000 per pay period. What is the mean pay per period in the office? What is the median?
......!!!!!!!!...................................
RESPONSE --> 10+5+2=17 10*700+5*800+2*1000=13,000 13,000/17=823is the mean 700 is the median confidence assessment: 2
.................................................
......!!!!!!!!...................................
07:48:27 `q008. In the preceding problem ten employees make $700 per pay period, while five make $800 per pay period and the other two make $1000 per pay period; we just found that the mean pay per period was $823. On the average, how much to the individual salaries differ from the mean?
......!!!!!!!!...................................
RESPONSE --> The mean is 823 823-700=123 823-800=23 1000-823=177 10*123+5*23+2*177=1630 1630/17=$96 confidence assessment: 2
.................................................
㝲P㢄~締 assignment #015 015. range vs ave dev vs std dev Liberal Arts Mathematics II 12-07-2008
......!!!!!!!!...................................
07:53:14 `q001. Note that there are 8 questions in this assignment. {}{}In what ways can you measure how 'spread out' the distribution 7, 9, 10, 11, 12, 14 is?
......!!!!!!!!...................................
RESPONSE --> 7+9+10+11+12+14/6=10.5 the deviation from the mean is 3.5+2.5+.5+.5+1.5+3.5/6=2 14-7=7 This is the range confidence assessment: 2
.................................................
......!!!!!!!!...................................
07:53:57 `q002. Comparing the distributions 7, 9, 10, 11, 12, 14 and 7, 8, 9, 12, 13, 14, which distribution would you say is more spread out?
......!!!!!!!!...................................
RESPONSE --> both groups have the same range confidence assessment: 3
.................................................
......!!!!!!!!...................................
10:57:25 `q003. Another measure of the spread of a distribution is what is called the standard deviation. This quantity is similar in many respects to the average deviation, but this measure of deviation is more appropriate to statistical analysis. To calculate the standard deviation of a distribution of numbers, we begin as before by calculating the mean of the distribution and then use the mean to calculate the deviation of each number from the mean. For the distribution 7, 9, 10, 11, 12, 14 we found that the mean was to 10.5 deviations were 3.5, 2.5, .5, .5, 1.5 and 3.5. To calculate the standard deviation, we first square the deviations to find the squared deviations. We then average the squared deviations. What the you get for the squared deviations, then for the average of the squared deviations, for the given distribution?
......!!!!!!!!...................................
RESPONSE --> 3.5^2=12.25 2.5^2=6.25 5^2=.25 1.5^2=2.25 12.25+6.25+0.25+0.25+2.25+12.25/6=5.67 confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:57:49 `q004. In the last problem we calculated the average of the squared deviations. Since this average was calculated from the squared deviations, it seems appropriate to now take the square root of our result. The standard deviation is the square root of the average of the squared deviations. Continuing the last problem, what is the standard deviation of the distribution 7, 9, 10, 11, 12, 14?
......!!!!!!!!...................................
RESPONSE --> 2.4 confidence assessment: 1
.................................................
......!!!!!!!!...................................
12:29:18 `q005. The last problem didn't really lie to you, there is one more subtlety in the calculation of the standard deviation. When we calculate the standard deviation for a distribution containing less than about 30 numbers, then in the step where we calculated the average deviation we do something a little bit weird. Instead of dividing the total of the squared deviations by the number of values we totaled, we divide by 1 less than this number. So instead of dividing (12.25 + 6.25 + 0.25 + 0.25 + 2.25 + 12.25) by 6, as we did, we only divide by 5. With this modification, what is the standard deviation?
......!!!!!!!!...................................
RESPONSE --> (12.25.+6.25+0.25+2.25+12.25)=3.4 (12.25+6.25+0.25+0.25+2.25+12.25/5=6.8 2.6 confidence assessment: 2
.................................................
......!!!!!!!!...................................
12:38:33 `q006. We just calculated the standard deviation of the distribution 7, 9, 10, 11, 12, 14. Earlier we noted that the distribution 7, 8, 9, 12, 13, 14 is a bit more spread out than the distribution 7, 9, 10, 11, 12, 14. Calculate the standard deviation of the distribution 7, 8, 9, 12, 13, 14 and determine how much difference the greater spread makes in the standard deviation.
......!!!!!!!!...................................
RESPONSE --> The mean of the destribution is 7,8,9,12,13,14 = 10.5 the deviations are 3.5,2.5,1.5,1.5,2.5,and 3.5 the squared deviations are 12.25,6.25,2.25,2.25,6.25, and 12.25 =42 the avg is 42/5=8.4 std dev 1sqrt(8.4)=2.9 confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:46:50 `q007. What is the standard deviation of the distribution 7, 8, 8, 8, 13, 13, 13, 14. What would be the quickest way to calculate this standard deviation?
......!!!!!!!!...................................
RESPONSE --> The mean is 10.5 the deviations from the mean are 3.5, 2.5, 2.5,2.5,2.5, 2.5, 2.5, 3.5 the squared deviations are 12.25, 6.25, 6.25,6.25, 6.25,6.25, 6.25, 12.25 sum is 64 confidence assessment: 2
.................................................
......!!!!!!!!...................................
12:51:06 `q008. What is the maximum possible standard deviation for a set of six numbers ranging from 7 through 14 and averaging (7 + 14 ) / 2 = 10.5?
......!!!!!!!!...................................
RESPONSE --> 7,7,7,14,14,14 The deviation of the six #'s is 3.5 the mean is 10.5 squared =12.25 6*12.25=75 75/5=15=3.9 confidence assessment: 3
.................................................
You then start the process over with the next question, clicking on the Next Question/Solution button at top left, etc. &#