Normally distributed random variable?

Question

Hi.
I need help with the following as I can´t figure out how to get the answer. The textbook show the answer but I don´t know if it is right or how to get it.
Suppose x is a normally distributed random variable with u = 30 and o = 8. Find the random variable, call it a, such that:
P(x > a) = 0.1
Could you please explain each step so I can understand how this works?
How would I proceed if the problem was exactly the same except P(x < a) = 0.1
Thank you.

Answer 1

Nobody integrates this function. The probabilities are worked out for you in the standard normal table.
If µ = 30 and σ = 8, then to use the table, you must use the transformation:
Z = (x-µ)/σ, For example, given µ = 30 and σ = 8, find the probability that a random variable, a, is less than 20.
Z = (20 - 30)/8 = - 1.25 Go to the standard normal table and find the probability associated with Z = -1.25. The notation is:
Φ[-1.25] = 0.1056.
Example (2) Find the probability the random variable is greater than a = 45.
First find the probability that it is less than 45:
Z = (45-30)/8 = 1.875
Φ[1.875] = 0.9695
To find the probability above 45, subtract this from 1 (because the total probability under all probability distributions is equal to 1).
So the answer is 1 - 0.9695 = 0.0305.
Conversely, if you want to find the value of a for a probability of 0.10,
Then let Φ[Z] = 0.10. Find the probability in the table = 0.10 and work backwards to find the value of x
In this case, Z = -1.28 = (x-µ)/σ and solve for x.
x = -1.28(8) + 30 = 19.76

Answer 2

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RE:
Normally distributed random variable?
Hi.
I need help with the following as I can´t figure out how to get the answer. The textbook show the answer but I don´t know if it is right or how to get it.
Suppose x is a normally distributed random variable with u = 30 and o = 8. Find the random variable, call it a, such that:
P(x > a) =...

Answer 3

The probability density function f for a normal distribution is

f(x) = 1/(sigma sqrt(2 pi)) exp[-(x - mu)^2/(2 sigma^2)]

When you integrate this from a to infinity you get the probability that a random x is bigger than a. That is P(x > a). So

P(x > a) = integral f(x)dx

with the boundaries infinity and a. Solve this for a and substitute the values.

Answer 4

I am not familiar with your notation; is "u" the mean and "o" the sq rt of variance?

In any case, the value of x such that P(x>a) is the area of the probability curve from x = a to ∞.

0.1 = ∫[a->∞]p(x)dx, where p(x) is the probability density function for the normal distribution.

For P(x
0.1 = ∫[-∞->a]p(x)dx

p(x) = (1/2πo)e^ -{[(x-u)^2]/so^2}

The integral cannot be expressed in closed form, but is expressed in terms of the erf function:

∫[-∞->a]p(x)dx = (1/2)[1 + erf{(x-u)/o√2}]

and

∫[a->∞]p(x)dx = 1 - ∫[-∞->a]p(x)dx