Question

Assume X is normally distributed with a mean of 9 and a standard deviation of 2. Determine the value for x that solves each of the following. Round the answers to 2 decimal places. a) P(X > x) = 0.5. b) P(X > x) = 0.95. x= c) P(x < X < 9) = 0.2. x = i d) P(-x< X - 9 < x) = 0.95. x= i e) P(-x< X - 9 < x) = 0.99. x= i

Answer 1

Solution:

Given, the Normal distribution with,

  $\mu$ = 9

$\sigma$ = 2

a)P(X > x) = 0.5

P(X < x) = 1 - 0.5

P(X < x) = 0.5

For z ,

P(Z < z ) = 0.5

From z table , P(Z < 0) = 0.5

So , z = 0

Using z score formula , x = $\mu$ + z$\sigma$ = 9 + (0 * 2 ) = 9.00

Answer: x = 9.00

b)

P(X > x) = 0.95

P(X < x) = 1 - 0.95

P(X < x) = 0.05

For z ,

P(Z < z ) = 0.05

From z table , P(Z < -1.645) = 0.5

So , z = -1.645

Using z score formula , x = $\mu$ + z$\sigma$ = 9 + (-1.645 * 2 ) = 5.71

Answer: x = 5.71

c) P(x < X < 9) = 0.2

P(X < 9) - P(X < x) = 0.2

P[(X - $url=https://latex.codecogs.com/gif.latex?%5Cmu&t=1572452490&ymreqid=50870fb5-0ab3-84d7-1ca5-8f000101df00&sig=MPNQBgISe9gG1ditYIG8XA--~C$ )/$url=https://latex.codecogs.com/gif.latex?%5Csigma&t=1572452490&ymreqid=50870fb5-0ab3-84d7-1ca5-8f000101df00&sig=FimOGlwrVkizfR_6z6asyQ--~C$ <  (9 - 9)/2] - P(X < x) = 0.2

P[Z < 0.00] - P(X < x) = 0.2

0.5 - P(X < x) = 0.2

P(X < x) = 0.3

For z , P(Z < z ) = 0.3

From z table , P(Z < -0.524 ) = 0.3

So, z = -0.524

Using z score formula , x = $\mu$ + z$\sigma$ = 9 + (-0.524 * 2 ) = 7.95

x = 7.95

d)P(-x < X - 9 < x) = 0.95

Since normal distribution is symmetric ,

P[(X - 9) < -x] = 0.025 and P[(X - 9) > x] = 0.025

Consider , P[(X - 9) < -x] = 0.025

$\therefore$ P[(X - 9)/$\sigma$ < -x/$\sigma$ ] = 0.025

$\therefore$ P[Z < -x/$\sigma$ ] = 0.025

But from z table , P(Z < -1.96 ] = 0.025

So we can write , -x/$\sigma$ = -1.96

$\therefore$  -x =  -1.645 * $\sigma$ =  -1.96* 2 = -3.92

$\therefore$x = 3.92

e)

P(-x < X - 9 < x) = 0.99

Since normal distribution is symmetric ,

P[(X - 9) < -x] = 0.005 and P[(X - 9) > x] = 0.005

Consider , P[(X - 9) < -x] = 0.005

$\therefore$ P[(X - 9)/$\sigma$ < -x/$\sigma$ ] = 0.005

$\therefore$ P[Z < -x/$\sigma$ ] = 0.005

But from z table , P(Z < -2.576 ] = 0.005

So we can write , -x/$\sigma$ =  -2.576

$\therefore$  -x =  -2.326 * $\sigma$ =  -2.576* 2 = -5.15

$\therefore$x = 5.15

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