6.2 Examples/Explanations
Question #1
P(2) = 40C2 (0.03)2 (1 – 0.03)(40 – 2) = 0.2206 when rounded to four places
Or just use “binompdf” on your calculator:
Question #2
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Meaning that we are being asked for the probability of 3 or fewer successes. So we will have to find all
four of these values.
P(0) = 9C0 (0.2)0 (1 – 0.2)(9 – 0)  0.134217728 (I will actually do the “binompdf” on calc)
P(1) = 9C1 (0.2)1 (1 – 0.2)(9 – 1)  0.301989888
P(2) = 9C2 (0.2)2 (1 – 0.2)(9 – 2)  0.301989888
P(3) = 9C3 (0.2)3 (1 – 0.2)(9 – 3)  0.176160768
Find the sum:
P(0) + P(1) + P(2) + P(3) = 0.134217728 + 0.301989888 +0.301989888 + 0.176160768
 0.9144 (when rounded to four places)
OR to do this more quickly, use the “cumulative function”, binomcdf.
This will add up the probabilities from 0 to whatever number you put in for “x”.
So if you use x = 3 it will give you the sum of the probabilities from 0 to 3.
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Question #3
You can do all 10 calculations one by one, or you can find these more quickly using the calculator by
putting in the numbers 0 through 9 into L1. Then put the formula into L2.
We know that n = 9 and p = 0.65 the “x” values are in L1, so we have
P(x) = binompdf(9, 0.65, L1)
So I will type this into L2 by going up to the top of the list where it says “L2” and hitting enter.
When you hit enter the calculator will display all the probabilites in L2.
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scroll down to see the rest
Round these to four decimal places to get your answers below:
np = 9(0.65)
= 5.85
Square root (np(1-p)) = square root (9*0.65 ( 1 – 0.65) )
= square root (5.85 * 0.35)
= 1.4 (rounded to one place)
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A histogram is skewed right if the tail to the right of the peak is longer than the tail to the left of the
peak. Conversely, a histogram is skewed left if the tail to the left of the peak is longer than the tail to
the right of the peak. It is symmetric if the length of both tails to the left and right are approximately
equal.
Please see page 4 of this pdf (click on link) for further instructions on how to do a binomial histogram on
your ti-84
Match the window up to your choices above
make sure only plot 1 is on
choose the third option (histogram) and make sure Freq is set to L2
make sure nothing is in Y1 or any other lines
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Hit “Graph” and you see this:
which clearly matches choice C most closely
Question #4
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1) There are a fixed number of trials (17 flights selected).
2) Each flight is independent because they were randomly selected and no flight’s
arrival depends on any other flight.
3) There are only two mutually exclusive outcomes – the flight is on time or it is not.
4) The probability of success (being on time) is the same (90%) for each trial (flight).
Therefore this is a binomial experiment.
n = 17 p = 0.90 x = 15
So, P(15) = 0.2800 when you round to four places
c) “at least” 15 means 15 or more. So that means the “complement” of 0 through 14 flights
being on time. So we can do 1 – P(14 or less) and get “15 or more”. To do this we will use
binomcdf since it will give us the sum of the probabilities from 0 to 14
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n = 17 p = 0.90 x = 15
P(x > 15) = 0.7618 (when rounded to four places)
Fewer than 15 would mean 0 through 14. We can use the binomial cdf function where x = 14.
rounded to four places P(x<15) = 0.2382
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You can either add up P(14) + P(15) + P(16) or you could find the probability of 16 or fewer
flights and subtract the probability of 13 or fewer flights.
Either way, your result will be 0.7506 when rounded to four places.
OR
Question #5
N = 50 p = 0.62 x = 39
round to four places
P(39) = 0.0071
Between 35 and 37 can be found by adding P(35) + P(36) + P(37)
OR you can take
P(x < 37) – P(x < 34)
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If np(1-p) > 10, then the probability distribution will be approximately bell shaped and the
empirical rule can be used!! So if the observation is less than two standard deviations below
the mean, or greater than two standard deviations above the mean, then the result is
considered unusual.
Check the value of np(1 – p)
50(0.62) ( 1 – 0.62) = 11.78
Since 11.78 is greater than 10, we can assume this is approximately normally distributed.
Find the mean and standard deviation.
np = 50(0.62)
= 31
Square root (np(1-p)) = square root (50*0.62 ( 1 – 0.62) )
= square root (31 * 0.38)
= 3.4322….
Now find the upper and lower “limits” for what would be an unusual value.
The mean plus two standard deviations: 31 + 2(3.4322) = 37.8644
The mean minus two standard deviations: 31 – 2(3.4322) = 24.1356
Since 19 is less than 24.1356, YES it would be an unusual value.
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Question #6
a) n = 450 p = 0.49
np = 450(0.49)
= 220.5 rounded to nearest whole number is 221
Square root (np(1-p)) = square root (450*0.49 ( 1 – 0.49) )
= square root (220.5 * 0.51)
= 10.60448… rounded to the nearest tenth is 10.6
b) read it – hopefully that makes sense why the choice is A!
c) If np(1-p) > 10, then the probability distribution will be approximately bell shaped and the
empirical rule can be used!! So if the observation is less than two standard deviations below
the mean, or greater than two standard deviations above the mean, then the result is
considered unusual.
Check the value of np(1 – p)
450(0.49) ( 1 – 0.49) = 112.455
Since this is greater than 10, we can assume this is approximately normally distributed.
Now find the upper and lower “limits” for what would be an unusual value.
The mean plus two standard deviations: 221 + 2(10.6) = 242.2 (anything larger than this would be
“unusual”)
The mean minus two standard deviations: 221 – 2(10.6) = 199.8 (anything smaller than this would be
“unusual”)
Since 229 is not less than 199.8 or bigger than 242.2, it would not be an unusual value.
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