Questions on Algebra: Probability and statistics answered by real tutors!

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1. Suppose a contest offers $1,000 to the person who can guess the winning four digit number. How many possibilities are there?
----------
# of four-digit numbers: 1000-9999= 8999+1 = 9000
-------------------------------------------------------

2. There are 10 women and 8 men in a club. How many different committees of 6 people can be selected from the group if equal numbers of men and women are to be on the committee?
----
6 on the committe requires 3 men and 3 women.
# of groups of 3 women: 10C3 = [10*9*8]/[1*2*3] = 120
# of groups of 3 men : 8C3 = 8*7*6/1*2*3 = 56
------------------
# of possible committees: 120*56 = 6720
============================================
Cheers,
Stan H.

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1. Find the number of permutations in the word GEOMETRY.
8!/2! = 20160 arrangements
-------------------------------------
2. In how many ways can 12 books be displayed on a shelf if 12 books are available.
12! = 479,001,600 ways
=============================
Cheers,
Stan H.
===============================

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The records of a casualty insurance company show that, in the past, its clients have had a mean of auto accidents per day with a variance of 1.8 . The actuaries of the company claim that the variance of 0.0016 the number of accidents per day is no longer equal to 0.0016 . Suppose that we want to carry out a hypothesis test to see if there is support for the actuaries' claim. State the null hypothesis Ho and the alternative hypothesis Hi that we would use for this test.
-----------
Ho: sigma^2 = 1.8
Ha: sigma^2 is not equal to 1.8
=====================================
Cheers,
Stan H.

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6 blue
8 red
5 purple
9 green
2 clear
---
30 total
---
probability of getting a clear marble on the first pick is 2/30 because there are 2 clear marbles out of a total of 30 marbles in the box.
---
probability of getting a purple marble on the second pick regardless of what was chosen on the first pick is 5/30 because there are 5 purple marbles out of a total of 30 marbles in the box once again since you put the marble you took on the first pick back in the box.
---
the probability of getting a clear marble on the first pick, putting it in the box, and then getting a purple marble on the second pick is 2/30 * 5/30.
---
if the probability of an event happening is x, and the probability of another event happening is y, then the probability of both events happening is x*y.
---
by replacing the clear marble in the box, you made the two individual events independent of each other. if you did NOT replace the clear marble, then the probability of getting a purple marble on the second pick would have changed.
in that case it would have been 5/29 rather than 5/30.
---

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Take note that there are 30 marbles (6 blue + 8 red + 5 purple + 9 green + 2 clear = 30 total)

P(picking clear marble on first pick) = # of clear marbles/# total = 2/30 = 1/15


Since the first marble was replaced, this means that


P(picking purple marble on second pick) = # of purple marbles/# total = 5/30 = 1/6



----------------------------------------------------


Now the chances of the two events occurring are:


P(picking clear marble on first pick AND picking purple marble on second pick) = P(picking clear marble on first pick) * P(picking purple marble on second pick) = (1/15)(1/6)=1/90


So the probability is 1/90

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Total number of marbles=30
2/30*5/30=10/900=1/90=.0111 or 1.11% success.

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It's not that easy drawing a tree diagram.
It's much easier with paper and pencil.
I've shown the complete tree for the first roll of a 1.
Then the second roll could be 1-6.
The sum is then also shown.
The complete tree diagram would also show similar trees for the second roll for 2-6.
I'll leave that for you to complete in addition to the sums.

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A.) The amount of money the CEO is paid.

Discrete. CEO cannot be paid less than a penny.


B.) The number of staff members who report to the CEO.

Discrete. Cannot have a fraction of a staff member.

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I would have said
a)qualitative - this person is giving a 'feeling' about something based on how dirty the source for energy is
b)quantitative - this is a hard and fast number

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This is a "goodness-of-fit-test".
There are 5 categories so df=5-1=4
-----------
With alpha = 1%, the critical Chi-square value is 13.277
======================
Cheers,
Stan H.

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Expected value = ](row total)/(grand total)] = 22.5
--------
Ans: C
===========
Cheers,
Stan H.

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It appears that the table was a 2 x 4 contingency table.
-----
the degrees of freedom would be (2-1)(4-1) = 3
-----
Usins a Chi-Sq test on the data with alpha = 1% the critical
Chi-square value would be 11.345 or Ans. C.
=================
Cheers,
Stan H.

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One group of accounting students took a distance learning class, while another group took the same course in a traditional classroom. At α = .10, is there a significant difference in the mean scores listed below?
(a) State the hypotheses.
Ho: u1 = u2
Ha: u1 is not equal to u2
---------------------------------
(b) State the decision rule and sketch it.
Critical value for 2-tail Z-test with alpha=0.10 is +/-1.645
--------------------------------
(c) Find the test statistic.
z(9.1-10.3) = -1.5485
(d) Make a decision.
Fail to Reject Ho.
(e) Use Excel to find the p-value and interpret it.
p-value = 0.1215
At least 12.15% of test results could have provided stronger
evidence for rejecting Ho.
===============================
Note: I used a 2-Sample ZTest on a TI-83 calculator to get these results.
===============================
Cheers,
Stan H.
Exam Scores for Accounting Students
Statistic Distance Classroom
Mean scores ¯x1 = 9.1 ¯x2 = 10.3
Sample std. dev. s1 = 2.4 s2 = 2.5
Number of students n1 = 20 n2 = 20

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need a problem???thanks

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Recently, students in a marketing research class were interested in the driving behavior of students. Specifically, the marketing students were interested if exceeding the speed limit was related to gender. They collected the following responses from 100 randomly selected students: Speeds Does not speed
Males 40 25
Females 10 25
1. The appropriate test to analyze the relationship between gender and education is:
A. regression analysis
B. Analysis of variance
C. Contingency table analysis
D. Goodness-of-fit
----
Ans: B
---------------------
2. The null hypothesis for the analysis is:
A. There is no relationship between gender and speeding.
B. The correlation between gender and speeding is zero.
C. As gender increases, speeding increases.
D. The mean of gender equals the mean of speeding.
---
Ans: A
-------------------
3. The degrees of freedom for the analysis is:
A. 1
B. 2
C. 3
D. 4
---
Ans: A
---------------------
4. Using 0.05 as the significance level, what is the critical value for the test statistic?
A. 9.488
B. 5.991
C. 7.815
D. 3.841
---
Ans: D
----------------------

5. What is the value of the test statistic?
A. 100
B. 9.89
C. 50
D. 4.94
---
Ans: B
--------------------------

6. Based on the analysis, what can be concluded?
A. Gender and speeding are correlated.
B. Gender and speeding are not related.
C. Gender and speeding are related.
D. No conclusion is possible.
---
Ans: A
-----------------------------
Cheers,
Stan H.

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In establishing warranties on HDTV sets, the manufacturer wants to set the limits so that few will need repair at manufacturer expense. On the other hand, the warranty period must be long enough to make the purchase attractive to the buyer. For a new HDTV the mean number of months until repairs are needed is 36.84 with a standard deviation of 3.34 months. Where should the warranty limits be set so that only 10 percent of the HDTVs need repairs at the manufacturer's expense?
--------------------------------
Draw a normal curve with mean 36.84.
Sketch a left tail containing 10% of the population of HDTV that fail.
----------------
Find the z-value corresponding to left-tail 10%: -1.2816
-----------------
Find the x-value corresponding to that z-value:
z(x) = (x-u)/sigma
-1.2816 = (x-36.84)/3.34
x-36.84 = -4,2804
x = 32.56 months is the upper limit of time for the warranty.
==============================
Cheers,
Stan H.

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3t^2 + 5t - 7 divided by t-5
---
t goes into 3t^2 by a factor of 3t.
3t * (t-5) = 3t^2 - 15t
---
3t^2 + 5t - 7 minus 3t^2 - 15t equals 20t - 7
---
t goes into 20t by a factor of 20
20 * (t-5) = 20t - 100
---
20t - 7 minus 20t - 100 equals 93
---
the remainder is 93/(t-5)
---
the complete answer is 3t + 20 + 93/(t-5)
---
to prove this is true, then (t-5) * (3t + 20 + 93/(t-5)) should equal 3t^2 + 5t - 7
---
when you multiply polynomials, each factor in one polynomial has to be multiplied by each factor in the other polynomial.
this means that:
(a + b) * (c + d + e) equals (ac + ad + ae) + (bc + bd + be) = first way
alternatively, this also means that:
(a + b) * (c + d + e) equals (ac + bc) + (ad + bd) + (ae + be) = second way
---
the answer is the same either way.
---
multiplying them out the first way, you get:
(t-5) * (3t + 20 + 93/(t-5)) = t*(3t + 20 + 93/(t-5)) - 5*(3t + 20 + 93/(t-5))
which equals
3t^2 + 20t + 93t/(t-5) - 15t - 100 - 5*93/(t-5)
which equals
3t^2 + 5t - 100 + 93t/(t-5) - 5*93/(t-5)
which equals
3t^2 + 5t - 100 + 93*(t-5)/(t-5)
which equals
3t^2 + 5t - 100 + 93
which equals
3t^2 + 5t - 7
---
multiplying them out the second way, you get:
(t-5) * (3t + 20 + 93/(t-5)) = (t-5) * 3t) + ((t-5) * 20) + (t-5)*93/(t-5)
which equals
3t^2 - 15t + 20t - 100 + 93
which equals
3t^2 + 5t - 7
---
the answer is the same.
the remainder is 93/(t-5)
---

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the formula for the nth term of a geometric sequence is:
---

---
you are given and .
you use them to find r as follows:
which becomes
substituting 4 for and 2 for you get:
4 = 2*r^1 = 2*r
solving for r, you get
r = 2
---
you now have r, and you can substitute in the general equation as follows:

substitute 2 for r, and 2 for and you get:
= 2*2^9 = 2*512 = 1024.
---
the 10th term in the sequence is 1024.
you can test this by doing it the hard way (each term is done separately) as follows:
a(1) = 2
a(2) = 4
a(3) = 8
a(4) = 16
a(5) = 32
a(6) = 64
a(7) = 128
a(8) = 256
a(9) = 512
a(10) = 1024
---
each succeeding number is multiplied by the ratio.
you do that 9 times which is the same as multiplying the original number by 2^9.

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The first two terms of a geometric sequence are and . How do I find , the eighth term?

First find the common ratio, , by dividing any term by
its preceding term.  So 



Now use the formula for the th term:

 

Substitute 








            




Edwin

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You want the sum of the integers 1 through 10.

The first number in your series is 1 and the last number in your series is 10.



The second number in your series is 2 and the second to the last is 9.



Since you have 10 numbers total, you have half that many, or 5 such pairs that add up to 11.



This works in general. If you want the sum of a series of integers, add the first number to the last number, multiply that sum by the number of numbers in the series, and then divide by 2. Symbolically:



where is the first number, is the last number, and is the number of numbers in the series.

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Sum = n/2 (t1 + t2) where n = total no. of terms, t1 = first term, t2 = last term
:
remember zero is neither positive or negative

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        b2   -3b  +18
      --------------------- 
b+7 ) b3 + 4b2 - 3b + 126
      b3 +7b2
        ----------------
          -3b2 - 3b
          -3b2 -21b
             ---------
                18b+126
                18b+126
                --------
                      0
Answer: b2 -3b  +18 
 

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5 CHEVROLETS ARE CRASH TESTED TO FIND OUT THEIR MEAN REPAIR COST. IT WAS FOUND THAT THE MEAN REPAIR COST OF THIS SMALL SAMPLE WAS $1706 WITH AN S.D OF $830. cONSTRUCT A 90% CONFIDENCE INTERVAL FOR THE MEAN REPAIR COST IN ALL SUCH VEHICLE COLLISIONS.
------------------------------------------
x-bar = 1706
E = t(10% for a 2-tailed test with df=4)[830/sqrt(5)]
= 1.533[371.19] = 569.03
--------------------
90% CI: 1706-569.03 < u < 1706+569.03
==============================================
Cheers,
Stan H.

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The standard deviation number is missing.
Cheers,
Stan H.

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An exeriment to measure the effect of giving growth hormones to girls affected by Turner's Syndrome was carried out recently in Vancouver. All 34 girls in the study were given the growth hormone and their heights were measured at the time the hormone was given and again one year later. No measurements were made on their final adult heights. Which of the following is not a problem with this experiment:
(a) There was no blinding.
(b) There was no control group.
(c) Nonresponse bias.
(d) There was insufficient attentionto the placebo effect.
(e) Because final heights were not measured, it is impossible to tell if the hormone affected final height or only accelerated growth and made no difference to final height.
------------------
(d) is not a problem. There is no possible placebo effect where measurements are involved.

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what number should be added to the following set of data so that the mean,median,and mode will become the same number ?
91,93,93,95,95,98,100

Since there are two 93's and two 95's, it order to have 
a mode (the number that occurs the most number of times),
the number to add must either be a 93 or a 95.  The 95
is right in the middle, and the 93 is more toward the 
bottoms, so 95 seems the most promising, so I'll add it
in the middle and check to see if it works:

91,93,93,95,95,95,98,100

91+93+93+95+95+95+98+100=760

760÷8 = 95, the mean.

The middle two numbers are both 95, so 95 is the median.

There are three 95's so there are more of them than
any other number, so 95 is the mode, too.

Edwin

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91,93,93,(95),95,95,98,100
Proof:
Sum=760
760/8=95 median, mode, mean.

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Do you still need help with these problems?
I see they were posted a week ago.
Cheers,
Stan H.

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done 12/23
Cheers,
Stan H.

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"Nonparametric tests are sometimes called distribution free statistics because they do not require that the data fit a normal distribution."
------------------------------------------------------------------
Comment: You have listed your answers. I will append what I think
are the answers. But your opinion might very well be as good as
mine on some answers.
--------------
Label each of the following situations “P” if it is an example of parametric data or “NP” if it is an example of nonparametric data.
1. A manufacturer produces a batch of memory chips (RAM) and measures the mean-time-between-failures (MTBF). The manufacturer then changes a manufacturing process and produces another batch and again measures the MTBF. Did the change to the process improve the MTBF? _NP_ I agree
-------------------------------------------------
2. From a written survey where the respondents were asked to rate an individual on a scale of 1 to 5, one group rated an individual a 3.7, another group rated the individual a 4.3. Is the difference statistically significant? __P__
If the rating scale is continuous the answer would be NP. If the scale is
discreet the answer would be P.
I believe the answer is NP but would say that the description is ambiguous.
------------------------------------------------
3. A catering company is buying equipment in order to set up their own store. They have a choice of two ovens that they can purchase for the store. The used oven is $100 less than the new oven, but its heating calibration is off by 20 degrees. Which one is a better buy for them? ____
I see no way of performing this test with parameters. I would say NP.
--------------------------------------------------
4. Jim Smith owns three real estate offices in Anytown. He has decided to open one more office, but he cannot decide between Hometown or Uptown as the town where he wants to locate. He will be comparing the mean number of homes sold per real estate agent, and the mean commission percentage earned by agents in the two towns to make his decision. _P_ I agree
---------------------------------------------------
5. A study to determine if job absenteeism is distributed evenly over the week. _NP_
I think it would be a parametric test with P(monday)=P(tuesday)...=P(friday)
I say "P".
----------------------------------------------------
6. Mel’s Diner has been surveying their customers for the past couple of years about their dining experience in the restaurant. The survey uses a scale of one to five, five being best to indicate customer satisfaction. Mel’s customer satisfaction averaged 2.5 last year, but this year it is 2.9. Is this difference statistically significant? ____ I think "P".
---------------------------------------------------

7. Sally’s Beauty Salon just opened for business. Sally assigns the stylists customers on a rotation basis so that everyone is kept busy all day. One month after she opened the salon, Sally’s customer count for each stylist was (a) 20 customers; (b) 30 customers; (c) 15 customers; and (d) 25 customers. Has Sally been fair in how she allocates customers to each of the stylists? ____
This would be a test of equal proportions. I say "P".
--------------------------------------------------------
8. A comparison of salaries between male and female employees in the same organization. _P__ I agree.
===============================
Cheers,
Stan H.

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Random sample of 120 customers’s spent an average of 9.2 hours on their professional job with a sample standard deviation of 5.1 hours. Calculate the specification hours with confidence of 95%.
-----------------------
x-bar = 9.2
E = 1.96[5.1/sqrt(120)] = 4.21
-------------------
95% CI: 9.2-4.21 < u < 9.2+4.21
==================================
Cheers,
Stan H.

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An auditor reviewed 25 oral surgery insurance claims from a particular surgical office, determining that the mean out-of-pocket patient billing above the reimbursed amount was $275.66 with a standard deviation of $78.11.
(a) At the 5 percent level of significance, does this sample prove a violation of the guideline that the average patient should pay no more than $250 out-of-pocket? State your hypotheses and decision rule.
---
Ho: u = 250
Ha: u > 250
----
Critical value for a one-tail test with alpha = 5%: z = 1.645
---
Test statistic: z(275.66) = (275.66-250)/[78.11/Sqrt(25)] = 1.6456
-------------------
Conclusion: Since the TS is not in the reject interval, Fail to reject Ho.
-------------------
(b) Is this a close decision?
Yes, the test statistic is very close to the critical value.
p-value = P(z > 1.6456) = 0.05024, which is very close to the alpha value.
There are only 5.024% of test results that could have provided stronger
evidence for rejecting Ho.
============================
Cheers,
Stan H.

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A coin was flipped 60 times and came up heads 38 times.
(a) At the .10 level of significance, is the coin biased toward heads? Show your decision rule and calculations.
----
Ho: p = 1/2
Ha: p is not 1/2
--------------
Critical value for 2-tail test with alpha = 1%: +/- 2.5758..
---------------
Test statistic: z(28/60) = [(28/60)-0.5]/[0.5*0.5/sqrt(60)] = -1.0328
------------------
Conclusion: Since TS is not in the rejection interval, Fail to Reject Ho.
The test provides evidence that the coin is not biased.
=================

(b) Calculate a p-value and interpret it.
P(z<-1.0328) = 0.150845
Since this is a 2-tail test the p-value is 2(0.150845) or appox. 30%
----------
Interpretation: Apporximately 30% of test results could have provided
stronger evidence for rejecting Ho, i.e. that the coin is biased.
===========================
Cheers,
Stan H.

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Faced with rising fax costs, a firm issued a guideline that transmission of 10 pages or more should be sent by 2-day mail instead. Exceptions are allowed, but they want the average to be 10 or below.
---------------
The firm examined 35 randomly chosen fax transmissions during the next year, yielding a sample mean of 14.44 with a standard deviation of 4.45 pages.
---------------
(a) At the .01 level of significance, is the true mean greater than 10?
Ho: u = 10
Ha: u > 10
Critical value for one-tail test with alpha = 1% = 2.32634
Test statistic: z(14.44) = (14.44-10)/[4.45/sqrt(35)] = 5.9027
--------
Conclusion: Since the test statistic is in the "Reject" interval,
reject Ho. The mean is greater than 10.
-------------------

(b) Use Excel to find the right-tail p-value.
p-value = P(z>5.9027) = 0.00000001793...
--------------------------------------------------
(a) At the .01 level of significance, is the true mean greater than 10?
The test provides evidence that the mean is not 10. Depending on how
your instructor interprets hypothesis tests, that is all the test shows.
Some instructors teach that rejecting Ho leads to accepting Ha: but not
all.
The test is a test of Ho, not a test of Ha.
===========
Cheers,
Stan H.

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A bag Cantains 6 White and 4 black balls, two
balls are drawm at random. Find the probability
that they are one of the same colour:-
--------------
P(2white or 2black) = P(2white) + P(2black) = 6C2/10C2 + 4C2/10C2
= (6C2 + 4C2)/(10C2) = (15 + 6)/45 = 21/45
=============================================
Cheers,
Stan H.

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The probability of rain is 20% on Friday,10% on Saturday,And 60% on Sunday.What is the probability that there is no rain on any of these three days?(To the nearest tenth of a percent).
------------
P(no rain on Friday) = 0.8
P(no rain on Saturday) = 0.9
P(no rain on Sunday) = 0.4
-------------------------------
Assuming the rain on each day is independent,
P(no rain on F or S or S) = 0.8*0.9*0.4 = 0.288 or 28.8%
===========================
Cheers,
Stan H.

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485/500=.97 or 97% which means they are exceeding their goals

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compute the geometric and harmonic means for the following distribution of annual death rates:
xi[3.95,4.95,5.95,6.95,7.95,8.95,9.95,10.95,11.95,12.95,13.95

You just have to learn some rules about the different kinds of averaging:

You didn't ask about arithmetic mean, but we have to use it
to find the harmonic mean.

How to find the arithmetic mean of some numbers:

1. Count them.
2. Add them all together.
3. Divide the result of 2 by the result of 1.

How to find the geometric mean of some numbers:

1. Count them.
2. Multiply them all together.
3. Take the root corresponding to 1 of the result of 1.

How to find the harmonic mean of some numbers:

1. Take the reciprocals of all the numbers
2. Take the arithmetic mean of the results of 1
3. Take the reciprocal of the result of 2.

3.95,4.95,5.95,6.95,7.95,8.95,9.95,10.95,11.95,12.95,13.95

To find the geometric mean:

1. Count them.

There are 11 of them

2. Multiply them all together.

(3.95)(4.95)(5.95)(6.95)(7.95)(8.95)(9.95)(10.95)(11.95)(12.95)(13.95)
= 1.353141797 × 1010.

3. Take the root corresponding to 1 of the result of 1.

Since we counted 11 in the first step, we take the 11th root:



----

To find the harmonic means:

1. Take the reciprocals of all the numbers:

,,,,,,,,,,

2. Take the arithmetic mean of the results of 1

To find the arithmetic mean of those numbers:

      1a. Count them.  There are 11
      2a. Add them all together. 1.19694482
      3a. Divide the result of 2 by the result of 1a.
           

3. Take the reciprocal of the result of 2.

       

Edwin

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Find the geometric mean of 45,32,37,46,39,36,41,48,36.

The geometric mean of N numbers is the Nth root of their
product.

So you want the 9th root of the product of those 9 numbers:



Edwin

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=39.68

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13*12*11*10*9=154,440 ans.

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35=(25+28-x)+2
35=53-x+2
x=53+2-35
x=55-25
x=20 number of programmers who know both languages.

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Since the combined total of those who either know
fortran or pascal is is greater than
the total number of programmers, there must be some
who know both. there are 2 who know neither
The total of those who know 1 or both languages is
therefore
Call the overlap of those who know both
25 fortran programmers include the overlap
28 Pascal programmers include the overlap also, so
I must subtract from



Twenty know both languages

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(a) Construct a 95 percent confidence interval for the true mean.

On your TI-83 or 84 calculator

Press STAT then ENTER

Put all 20 numbers in L1

Press STAT, press right arrow once to highlight CALC
Press ENTER twice.

You'll see some statistics come up on the screen

Press STAT then the right arrow twice to highlight TESTS
Then press 8 to get T-Interval

You see the TInterval menu.
Make sure Stats is highlighted

You should see:

TInterval
 Inpt:Data Stats
 x:346.5
 Sx:170.3783714...
 n:20
 C-Level:.95
 Calculate

Type .95 after C-Level if it's not there

Scroll down to highlight "Calculate"

Press ENTER

You should see 
TInterval
(266.76,426.24)
x=346.5
Sx=170.3783715
n=20

That's your confidence interval

(266.76,426.24)

or maybe your book writes it as 



--------------

.b) Why might normality be an issue here?

Because it is a small sample and we must assume
that the data be nomally distributed.

--------------

(c) What sample size would be needed to obtain an error of ±10 square millimeters with 99 percent confidence?

We use the formula:



First we find 

Subtract .99 from 1, getting .01, then divide this by 2
to get .005 

Press 2nd then VARS to get the DISTR menu
Press 3 to get invNorm( on the main screen

then after InvNorm(, type .005 and close the
parentheses, so that you see

InvNorm(.005)

on the main screen.

Press ENTER and you read -2.575829303

Now substitute in the formula:





Get 1926.030165

Round UP to 1927, (even though you were told in 
elementary school to round down when the first
digit dropped is less than 5) 
---------------

(d) If this is not a reasonable requirement,

This is not reasonable because there are only 1591 pages
in the directory.

suggest one that is.

Reduce the confidence level requirement to 90%.

Do the above using .9 instead of .99 and get
that the sample size needed is 786, which is not
unreasonable.

Edwin

You can put this solution on YOUR website!
1. Find the probability of 7 tomatos getting accepting in 6 stores when the mean is 5.5
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Mean of what? Do you have a standard deviation?
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2.Random sample of 120 customers’s spent an average of 9.2 hours on their professional job with a sample standard deviation of 5.1 hours. Calculate the specification hours with confidence of 95%.
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x-bar = 9.2 hrs
E = 1.96*(5.1/sqrt(120)) = 0.9125
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95% CI: 9.2 - 0.9125 < u < 9.2 + 0.9125
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3.In a set of 10 samples of bike, the mileage were listed as follows
41, 42, 45, 42, 47, 59, 55, 60, 40, 35
What percent of samples are getting more than an average of 40 ?
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# above 40 = 8
Total number = 10
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Percentage = 8/10 = 80%
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Cheers,
Stan H.