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$X$ is a normally distributed random variable with mean $44$ and standard deviation $13$ . What is the probability that $X$ is between $5$ and $83$ ? Use the $0.68-0.95-0.997$ rule and write your answer as a decimal. Round to the nearest thousandth if necessary.

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Find probabilities using the normal distribution I

Full solution

Q.

X

is a normally distributed random variable with mean

44

and standard deviation

13

. What is the probability that

X

is between

5

and

83

? Use the

0.68-0.95-0.997

rule and write your answer as a decimal. Round to the nearest thousandth if necessary.

Calculate z-score for $X=5$ : Mean ( $\mu$ ) is $44$ and standard deviation ( $\sigma$ ) is $13$ . Calculate the z-score for $X=5$ . $\newline$ $Z = \frac{X - \mu}{\sigma} = \frac{5 - 44}{13} = \frac{-39}{13} = -3$ .
Calculate z-score for $X=83$ : Calculate the z-score for $X=83$ . $Z = \frac{X - \mu}{\sigma} = \frac{83 - 44}{13} = \frac{39}{13} = 3$ .
Probability within standard deviations: Using the $0.68$ - $0.95$ - $0.997$ rule, the probability that $X$ is within $1$ standard deviation ( $\sigma$ ) of the mean ( $\mu$ ) is $0.68$ , within $2\sigma$ is $0.95$ , and within $0.95$ $0$ is $0.997$ .
Check z-scores within $3\sigma$ : Since the z-scores for $X=5$ and $X=83$ are $-3$ and $3$ , respectively, this falls within $3$ standard deviations from the mean.
Calculate probability between $5$ and $83$ : Therefore, the probability that $X$ is between $5$ and $83$ is approximately $0.997$ .

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