Suppose X is a normal random variable with $\mu = 40$ and $\sigma = 20$ . Find the value of $P ( 30 \leq X \leq 45 )$ . Please, round the answer to four decimal places.
A) $P ( 30 \leq X \leq 45 ) = 0.8549$
B) $P ( 30 \leq X \leq 45 ) = 0.2892$
C) $P ( 30 \leq X \leq 45 ) = 0.7098$
D) $P ( 30 \leq X \leq 45 ) = 0.2902$
E) $P ( 30 \leq X \leq 45 ) = 0.2912$

Question

Suppose  X is a normal random variable with $\mu = 40$ and $\sigma = 20$ . Find the value of $P ( 30 \leq X \leq 45 )$ . Please, round the answer to four decimal places.  
A) $P ( 30 \leq X \leq 45 ) = 0.8549$
B)   $P ( 30 \leq X \leq 45 ) = 0.2892$
C)   $P ( 30 \leq X \leq 45 ) = 0.7098$
D)   $P ( 30 \leq X \leq 45 ) = 0.2902$
E)   $P ( 30 \leq X \leq 45 ) = 0.2912$

Quizplus · Accepted Answer

To find $P(30 \leq X \leq 45)$, we convert the values to Z-scores using the formula $Z = \frac{X - \mu}{\sigma}$. For $X = 30$, $Z = \frac{30 - 40}{20} = -0.5$. For $X = 45$, $Z = \frac{45 - 40}{20} = 0.25$. Looking these Z-scores up in the standard normal distribution table or using a calculator, we find the probabilities for each and subtract the smaller from the larger to find the probability of $X$ being between 30 and 45. The correct probability is closest to the value given in option D, $0.2902$, after performing these calculations and considering rounding to four decimal places.

Suppose X Is a Normal Random Variable With μ=40\mu = 40μ=40

Suppose X Is a Normal Random Variable With $\mu = 40$