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$X$ is a normally distributed random variable with mean $29$ and standard deviation $11$ . What is the probability that $X$ is between $4$ and $54$ ? Write your answer as a decimal rounded to the nearest thousandth.

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Find probabilities using the normal distribution I

Full solution

Q.

X

is a normally distributed random variable with mean

29

and standard deviation

11

. What is the probability that

X

is between

4

and

54

? Write your answer as a decimal rounded to the nearest thousandth.

Calculate z-score for $X=4$ : Mean ( $\mu$ ) is $29$ and standard deviation ( $\sigma$ ) is $11$ . Calculate the z-score for $X=4$ . $\newline$ $Z = \frac{X - \mu}{\sigma} = \frac{4 - 29}{11} = \frac{-25}{11} \approx -2.273$
Calculate z-score for $X=54$ : Calculate the z-score for $X=54$ . $Z = \frac{X - \mu}{\sigma} = \frac{54 - 29}{11} = \frac{25}{11} \approx 2.273$
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 about $0.68$ , within $2\sigma$ is about $0.95$ , and within $3\sigma$ is about $0.997$ . $\newline$ Since our z-scores approximately equal $X$ $0$ and $X$ $1$ , this falls just beyond $X$ $2$ standard deviations from the mean.
Probability between $4$ and $54$ : The probability that $X$ is between $4$ and $54$ is slightly more than the probability of being within $2$ standard deviations, which is $0.95$ . We can estimate this probability to be slightly higher than $0.95$ .

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