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We delve deeper into the concepts introduced in vol.56, elucidating the process of determining the area (probability) of a normal distribution.
Properties of the normal distribution
The shape of a normal distribution is determined by the mean value (m) and standard deviation (σ) of the data. Figure 1 shows a normal distribution with mean value m = 60 points and standard deviation σ = 10 points. Let’s consider the properties of the normal distribution while analyzing the diagram.
[Figure 1]
The following points can be gleaned from Figure 1:
・It is symmetrical around the average value (60 points).
・The curve reaches its maximum at the average value and decreases as it spreads leftwards and rightwards.
・The points on the curve that correspond to the horizontal axes m-1σ (50 points in the figure) and m+1σ (70 points in the figure) are called “inflection points.” The part of the curve surrounded by the inflection point is convex upward, whereas the area outside the inflection point is convex downward.
・If the area surrounded by the curve and the horizontal axis is considered to be 100%, the area of the section within the curve is as shown in the table.
[Table]
How to find the area (probability) of a normal distribution
The area (probability) of a normal distribution can be calculated using an Excel function as follows.
When calculating the lower area (probability) of the horizontal axis value x in a normal distribution with mean value m and standard deviation σ:
＝NORM.DIST (x ,m,σ,1)
* 1 is a constant (1 is interpreted as TRUE.)
For m=60 , σ=10 , x=70
= NORM.DIST (70,60,10,1)
= 0.84
[Figure 2]
When calculating the horizontal axis value x for the lower surface area (probability) p:
=NORM.INV(p,m,σ)
For m=60, σ=10, p=0.84
=NORM.INV(0.84,60,10)
= 70
[Figure 3]
Utilizing the normal distribution
In the following example, the area (probability) of a normal distribution is calculated.
In mathematics, it is known that the deviation for 10,000 people who enter cram schools approximately follows a normal distribution. Determine how many deviation points are needed to fall within the range of 250.
Let A be the percentage of 10,000 people who are within the range of 250.
A=250÷10,000=0.025
Cumulative ratio = 1-0.025 = 0.975
[Figure 4]
Let x be the value on the horizontal axis where the ratio is 0.975. X is the desired score.
The average deviation value is 50 points, and the standard deviation is 10 points. The value of x for this normal distribution can be found as follows:
=NORM.INV(lower probability , mean value , standard deviation)
=NORM.INV(0.975,50,10)
= 69.6
From the above, students fall within the top 250 if they obtain a score of 70 or more.
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