A continuous random variable X has a normal distribution with mean 169. The probability that X takes a value greater than 180 is 0.17. Use this information and the symmetry of the density function to find the probability that X takes a value less than 158

Respuesta :

Answer:

0.17 is the probability that X takes a value less than 158.

Step-by-step explanation:

We are given the following in the question:

Mean = 169

The variable X is normally distributed.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

[tex]P(x > 180) = 0.17[/tex]

[tex]P( X > 180) = P( z > \displaystyle\frac{180 - 169}{\sigma})=0.17[/tex]  

[tex]= 1 -P( z \leq \displaystyle\frac{11}{\sigma})=0.17 [/tex]  

[tex]=P( z \leq \displaystyle\frac{11}{\sigma})=0.83 [/tex]  

Calculation the value from standard normal z table, we have,  

[tex]\displaystyle\frac{11}{\sigma} = 0.954\\\\\sigma = 11.53[/tex]  

We have to evaluate

P(x < 158)

[tex]P( x < 158) = P( z < \displaystyle\frac{158- 169}{11.53}) = P(z < -0.954)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(x < 158) = 0.170[/tex]

0.17 is the probability that X takes a value less than 158.