find the value of normal distribution with a 3 decimal places Z score using table
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In a past exam of my stats class, the question requires finding the value of a normal CDF corresponding to a Z score of 3 DP. However, we are only given a table that is accurate to 2 DP. We are not allowed to use calculator. What should I do to find the value when the Z score is 1.293, not just 1.29.
normal-distribution
asked Dec 29 '18 at 0:27
ZeyuanZeyuan
1273
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add a comment |
$begingroup$
In a past exam of my stats class, the question requires finding the value of a normal CDF corresponding to a Z score of 3 DP. However, we are only given a table that is accurate to 2 DP. We are not allowed to use calculator. What should I do to find the value when the Z score is 1.293, not just 1.29.
normal-distribution
asked Dec 29 '18 at 0:27
ZeyuanZeyuan
1273
$endgroup$
$begingroup$
You can linearly interpolate the values for $1.29$ and $1.3$
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– lulu
Dec 29 '18 at 0:34
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Would linear interpolation be accurate enough? If so using $1.29, 1.30$ and cutting three tenths between would be a possible way.
$endgroup$
– coffeemath
Dec 29 '18 at 0:35
add a comment |
$begingroup$
In a past exam of my stats class, the question requires finding the value of a normal CDF corresponding to a Z score of 3 DP. However, we are only given a table that is accurate to 2 DP. We are not allowed to use calculator. What should I do to find the value when the Z score is 1.293, not just 1.29.
normal-distribution
asked Dec 29 '18 at 0:27
ZeyuanZeyuan
1273
$endgroup$
In a past exam of my stats class, the question requires finding the value of a normal CDF corresponding to a Z score of 3 DP. However, we are only given a table that is accurate to 2 DP. We are not allowed to use calculator. What should I do to find the value when the Z score is 1.293, not just 1.29.
normal-distribution
normal-distribution
asked Dec 29 '18 at 0:27
ZeyuanZeyuan
1273
asked Dec 29 '18 at 0:27
ZeyuanZeyuan
1273
asked Dec 29 '18 at 0:27
ZeyuanZeyuan
1273
asked Dec 29 '18 at 0:27
ZeyuanZeyuan
1273
asked Dec 29 '18 at 0:27
ZeyuanZeyuan
1273
1273
$begingroup$
You can linearly interpolate the values for $1.29$ and $1.3$
$endgroup$
– lulu
Dec 29 '18 at 0:34
$begingroup$
Would linear interpolation be accurate enough? If so using $1.29, 1.30$ and cutting three tenths between would be a possible way.
$endgroup$
– coffeemath
Dec 29 '18 at 0:35
add a comment |
$begingroup$
You can linearly interpolate the values for $1.29$ and $1.3$
$endgroup$
– lulu
Dec 29 '18 at 0:34
$begingroup$
Would linear interpolation be accurate enough? If so using $1.29, 1.30$ and cutting three tenths between would be a possible way.
$endgroup$
– coffeemath
Dec 29 '18 at 0:35
$begingroup$
You can linearly interpolate the values for $1.29$ and $1.3$
$endgroup$
– lulu
Dec 29 '18 at 0:34
$begingroup$
You can linearly interpolate the values for $1.29$ and $1.3$
$endgroup$
– lulu
Dec 29 '18 at 0:34
$begingroup$
Would linear interpolation be accurate enough? If so using $1.29, 1.30$ and cutting three tenths between would be a possible way.
$endgroup$
– coffeemath
Dec 29 '18 at 0:35
$begingroup$
Would linear interpolation be accurate enough? If so using $1.29, 1.30$ and cutting three tenths between would be a possible way.
$endgroup$
– coffeemath
Dec 29 '18 at 0:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Personally, I would just round to $1.29$. However, if you really want to guess $text{normalcdf}(1.293)$, I would just do a weighted sum.
First, write $1.293$ in terms of $1.29$ and $1.30$, since those are the two-decimal z-scores closest to it:
$$1.293=0.7cdot 1.29+0.3cdot 1.30$$
Then, pretend $text{normalcdf}$ is linear:
$$text{normalcdf}(1.293)=0.7text{normalcdf}(1.29)+0.3text{normalcdf}(1.30) \ =0.7cdot 0.9015+0.3cdot 0.9032=0.90201$$
answered Dec 29 '18 at 0:36
Noble MushtakNoble Mushtak
15.3k1835
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add a comment |
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1 Answer
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$begingroup$
Personally, I would just round to $1.29$. However, if you really want to guess $text{normalcdf}(1.293)$, I would just do a weighted sum.
First, write $1.293$ in terms of $1.29$ and $1.30$, since those are the two-decimal z-scores closest to it:
$$1.293=0.7cdot 1.29+0.3cdot 1.30$$
Then, pretend $text{normalcdf}$ is linear:
$$text{normalcdf}(1.293)=0.7text{normalcdf}(1.29)+0.3text{normalcdf}(1.30) \ =0.7cdot 0.9015+0.3cdot 0.9032=0.90201$$
answered Dec 29 '18 at 0:36
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
Personally, I would just round to $1.29$. However, if you really want to guess $text{normalcdf}(1.293)$, I would just do a weighted sum.
First, write $1.293$ in terms of $1.29$ and $1.30$, since those are the two-decimal z-scores closest to it:
$$1.293=0.7cdot 1.29+0.3cdot 1.30$$
Then, pretend $text{normalcdf}$ is linear:
$$text{normalcdf}(1.293)=0.7text{normalcdf}(1.29)+0.3text{normalcdf}(1.30) \ =0.7cdot 0.9015+0.3cdot 0.9032=0.90201$$
answered Dec 29 '18 at 0:36
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
Personally, I would just round to $1.29$. However, if you really want to guess $text{normalcdf}(1.293)$, I would just do a weighted sum.
First, write $1.293$ in terms of $1.29$ and $1.30$, since those are the two-decimal z-scores closest to it:
$$1.293=0.7cdot 1.29+0.3cdot 1.30$$
Then, pretend $text{normalcdf}$ is linear:
$$text{normalcdf}(1.293)=0.7text{normalcdf}(1.29)+0.3text{normalcdf}(1.30) \ =0.7cdot 0.9015+0.3cdot 0.9032=0.90201$$
answered Dec 29 '18 at 0:36
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
Personally, I would just round to $1.29$. However, if you really want to guess $text{normalcdf}(1.293)$, I would just do a weighted sum.
First, write $1.293$ in terms of $1.29$ and $1.30$, since those are the two-decimal z-scores closest to it:
$$1.293=0.7cdot 1.29+0.3cdot 1.30$$
Then, pretend $text{normalcdf}$ is linear:
$$text{normalcdf}(1.293)=0.7text{normalcdf}(1.29)+0.3text{normalcdf}(1.30) \ =0.7cdot 0.9015+0.3cdot 0.9032=0.90201$$
answered Dec 29 '18 at 0:36
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 29 '18 at 0:36
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 29 '18 at 0:36
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 29 '18 at 0:36
Noble MushtakNoble Mushtak
15.3k1835
15.3k1835
add a comment |
add a comment |
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$begingroup$
You can linearly interpolate the values for $1.29$ and $1.3$
$endgroup$
– lulu
Dec 29 '18 at 0:34
$begingroup$
Would linear interpolation be accurate enough? If so using $1.29, 1.30$ and cutting three tenths between would be a possible way.
$endgroup$
– coffeemath
Dec 29 '18 at 0:35