Question

Let x be a continuous random variable that is normally distributed with a mean of 136.4 and a standard deviation of 18.5. Find the probability that x assumes a value between 109.1 and 130.5.Round your answer to four decimal places.The probability = Enter your answer in accordance to the question statement

Answer 1

Probability that x assumes a value between 109.1 and 130.5:

$P(109.11. Find the z-score for P(x>109.1): use the next formula:[tex]z=\frac{(x-\operatorname{mean})}{\text{standard deviation}}$$z=\frac{109.1-136.4}{18.5}=\frac{-27.3}{18.5}=-1.47$

Using a z-score table find the corresponding value for z=-1.47:

2. Find the z-score for P(x<130.5)

$z=\frac{130.5-136.4}{18.5}=\frac{-5.9}{18.5}=-0.32$

3. Subtract the z-score of upper limit by the z-score of lower limit:

$0.3745-0.0708=0.3037$

Then, the probability that x assumes a value between 109.1 and 130.5 is:

[tex]P(109.1