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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

User Starlight
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Answer:

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

Explanation:

We are given the following in the question:

Mean = 169

The variable X is normally distributed.

Formula:


z_(score) = \displaystyle(x-\mu)/(\sigma)


P(x > 180) = 0.17


P( X > 180) = P( z > \displaystyle(180 - 169)/(\sigma))=0.17


= 1 -P( z \leq \displaystyle(11)/(\sigma))=0.17


=P( z \leq \displaystyle(11)/(\sigma))=0.83

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


\displaystyle(11)/(\sigma) = 0.954\\\\\sigma = 11.53

We have to evaluate

P(x < 158)


P( x < 158) = P( z < \displaystyle(158- 169)/(11.53)) = P(z < -0.954)

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


P(x < 158) = 0.170

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

User Viswanathgs
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