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# Download For a normal distribution with mean 28 units and standard deviation

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```For a normal distribution with mean 28 units and standard deviation 3 units, to find
the area under THIS normal curve from
x30 to x35
fx 
2
2
1 e − x−28 / 23
3 2
fx
y
0.12
0.10
0.08
0.06
0.04
0.02
0.00
16
35

30
18
20
2
2
1 e − x−28 / 23
3 2
22
24
26
28
30
32
dx  0. 242 677 208 9
Normal Distribution, mean 28, standard deviation 3,
34
36
38
40
x
area under the normal curve from x30 to x35
2ndVARS gives
DISTR
Normal Distribution, mean 28, standard deviation 3,
area under the normal curve to the right of x32


32
2
2
1 e − x−28 / 23
3 2
dx  9. 121 121 973  10 −2 
0. 0912
Normal Distribution, mean 28, standard deviation 3,
area under THIS normal curve to the left of x29
29
2
2
1 e − x−28 / 23
− 3 2

dx  0. 630 558 659 8
For a normal distribution x with mean 28 and standard deviation 3, to find the
92nd percentile, that is the value of x such that the area under THIS normal curve to
the left of x is 0.92
x
− x−28 2 / 23 2
1
e

− 3 2
dx . 92, Solution is: x  32. 215 214 68
```
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