TI84_and_Excel

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Area under the Normal Curve with TI-84 and Excel
Example problem
The picture
TI-84 solution
𝑃(𝑧 < 1.37)
β€œFind the area under the normal
curve to the left of 𝑧 = 1.37.”
Answer:
Area = 0.9147
Probability = 0.9147 or 91.47%
2ND DISTR (on the VARS key)
(-) 1 2ND EE (on comma key) 9 9
(calculator’s way to say βˆ’βˆž)
, comma 1.37 )
right paren ENTER
𝑃(𝑧 > 1.37)
β€œFind the area under the normal
curve to the right of 𝑧 = 1.37.”
Answer:
Area = 0.0853
Probability = 0.0853 or 8.53%
𝑃(0.55 < 𝑧 < 1.85)
β€œFind the area under the normal
curve between 𝑧 = 0.55 and 𝑧 =
1.85.”
Answer: Area = 0.2590
Probability = 0.2590 or 25.90%
Excel solution
0.914657 =NORMSDIST(1.37)
=NORMSDIST(z-score) gives you
the area to the left of that z-score.
Use =NORM.S.DIST(z-score) in
Excel 2010+
0.085343451
=1-NORMSDIST(1.37)
2ND DISTR (on the VARS key)
1.37 , comma
1 2ND EE (on comma key) 9 9
(calculator’s way to say ∞)
) right paren ENTER
To get area to the RIGHT of a
certain z-score, tell Excel
to subtract 1 minus the
area to the left of that z-score.
0.259002912
=NORMSDIST(1.85)NORMSDIST(0.55)
To get the area between two zscores, tell Excel to subtract
The area to the left of the higher
z-score, minus the area to the left
of the lower z-score.
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Example problem
The picture
TI-84 solution
β€œFind the total area under the
normal curve.”
You should know this very basic
special fact, that the total area
under a probability distribution is
always exactly precisely = 1.
β€œWhat is the area under the right
half of the normal curve?” and
β€œWhat is the area under the left
half of the normal curve?”
You should know these very
special facts about the Normal
Distribution:
ο‚· it is symmetric,
ο‚· 𝑧 = 0 is in the middle,
ο‚· half of the area (0.5) is to
the left
ο‚· and half of the area (0.5)
is to the right.
Excel solution
=NORMSDIST(1E99)-NORMSDIST(1E99)
Use the same 1E99 for ∞ and the
same -1E99 for βˆ’βˆž in Excel as
you use in the TI-84.
=NORMSDIST(1E99)NORMSDIST(0)
=NORMSDIST(0)-NORMSDIST(1E99)
The calculator has rounding errors
in the 10th decimal position.
The correct answers are both
exactly precisely
0.5000000000000000000000000.
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Example problem
The picture
TI-84 solution
𝑃(𝑧 < βˆ’2.02)
β€œFind the area under the normal
curve to the left of 𝑧 = βˆ’2.02.
Excel solution
=NORMSDIST(-2.02)
Answer: Area = 0.0217
Probability = 0.0217 or 2.17%
Area to the left of βˆ’2.02 plus
area to the right of 2.02 totals the
entire area under the curve,
exactly precisely 1.000000000000.
𝑃(𝑧 > βˆ’2.02)
β€œFind the area under the normal
curve to the right of 𝑧 = βˆ’2.02.”
Answer: Area = 0.9783
Probability = 0.9783 or 97.83%
𝑃(βˆ’1.38 < 𝑧 < βˆ’0.38)
β€œFind the area under the normal
curve between 𝑧 = βˆ’1.38 and
𝑧 = βˆ’0.38.”
Answer: Area = 0.2682
Probability = 0.2682 or 26.82%
𝑃(βˆ’1.79 < 𝑧 < 2.20)
β€œFind the area under the normal
curve between 𝑧 = βˆ’1.79 and
𝑧 = 2.20.”
= 1 – NORMSDIST(-2.02)
Area to the right of a z-score:
subtract 1 minus the area to its
left.
=NORMSDIST(-0.38)NORMSDIST(-1.38)
Area between two z-scores:
Subtract area to the left of the
larger z-score minus the area to
the left of the smaller z-score.
=NORMSDIST(2.20)-NORMSDIST(1.79)
Answer: Area = 0.9494
Probability = 0.9494 or 94.94%
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Example problem
The picture
TI-84 solution
2ND DISTR 3:invNorm(area)
β€œWhat z-score has that much area
to the left of it?”
What z-scores bound the middle
50% of the area?
Think and draw:
50% = 0.5000 shaded in middle
1 - 0.5000 = 0.5000 not shaded
Half of that is 0.2500
0.2500 in the left tail
0.2500 in the right tail
And by symmetry, the right side
boundary is at z = positive 0.6745.
Check:
Answer:
𝑧 = βˆ’0.6745 to 𝑧 = 0.6745
Check can be done with
NORMSDIST:
=NORMSDIST(0.6745)NORMSDIST(-0.6745)
What z-scores bound the middle
68% of the area?
Solution:
Think and draw:
68% = 0.6800 shaded in middle
1 – 0.6800 = 0.3200 not shaded
Half of that is 0.1600
0.1600 in the left tail
0.1600 in the right tail
That’s the z-value to use in the
answer. We rounded to ±0.9945.
Answer:
𝑧 = βˆ’0.9945 to 𝑧 = 0.9945
Excel solution
=NORMSINV(0.25)
=NORMSINV(0.1600)
Check:
This is why the Empirical Rule says
β€œIn a Normal Distribution, about
68% of the data lies within 1
standard deviation of the mean.”
Yes, that rounds to 68%.
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Example problem
The picture
TI-84 solution
What z-scores bound the middle
95% of the area?
Think and draw:
95% = 0.9500 shaded in middle
1 – 0.9500 = 0.0500 not shaded
Half of that is 0.0250
0.0250 in the left tail
0.0250 in the right tail
Answer:
𝑧 = βˆ’1.9600 to 𝑧 = 1.9600
Or 𝑧 = βˆ’1.96 to 𝑧 = 1.96
What z-scores bound the middle
99.7% of the area?
Think and draw:
99.7% = 0.9970
1 – 0.9970 = 0.0030
Half of that is 0.0015
0.0015 in the left tail
0.0015 in the right tail
Answer:
𝑧 = βˆ’2.967 to 𝑧 = 2.967
Or 𝑧 = βˆ’2.97 to 𝑧 = 2.97
Excel solution
=NORMSINV(0.0250)
Rounded to 𝑧 = ±1.9600
Check:
This is why the Empirical Rule says
β€œIn a Normal Distribution, about
95% of the data lies within 2
standard deviations of the mean.”
=NORMSINV(0.0015)
Check:
This is why the Empirical Rule says
β€œIn a Normal Distribution, about
95% of the data lies within 2
standard deviations of the mean.”
Check Empircal Rule’s rounded
value of 𝑧 = ±3:
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Example problem
The picture
TI-84 solution
What z-score separates the top
10% of the data from the bottom
90%?
Excel solution
=NORMSINV(0.9000)
Check:
Think and draw:
90% = 0.9000 area is to the left.
10% = 0.1000 area is to the right
Answer:
𝑧 = 1.2816 or 𝑧 = 1.28
What score separates the bottom
third of the data from the top
two-thirds?
=NORMSINV(1/3)
Think and draw:
1/3 to the left. You can use the
fraction in TI-84 or in Excel.
Answer:
𝑧 = βˆ’0.4307 or 𝑧 = βˆ’0.43
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TI-84 Production notes
The TI-84 graphs were done as described below. Perhaps the ShadeNorm command should have been used instead. That’s something that
can be investigated another day. ShadeNorm documentation is shown on the next page.
The π‘₯-axis has the 𝑧 values from βˆ’4
to 4 with tickmarks each 𝑧 = 1 unit.
The 𝑦-axis goes from βˆ’0.1 to 0.4
with tickmarks every 0.1 unit.
The βˆ’0.1 was chosen to leave space
at the bottom of the screen.
For shading,
2ND CALC
Lower Limit and Upper Limit were
given and a value was displayed.
As discussed somewhere else, this
value is not always accurate
enough because the infinitely long
tails are missed with our choice of
finite Xmin and Xmax values.
So we clear the screen and press
GRAPH again to get the shaded
graph without the inadequate
∫ 𝑓(π‘₯) 𝑑π‘₯ result.
Then 2ND DRAW 1:ClrDraw clears
the shading in preparation for
another problem.
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TI-84 ShadeNorm command
Example: 2ND DISTR, DRAW submenu, 1:ShadeNorm
ShadeNorm(-1,1)
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