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$X$ is a normally distributed random variable with mean $93$ and standard deviation $1$ . What is the probability that $X$ is between $90$ and $96$ ? 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

93

and standard deviation

1

. What is the probability that

X

is between

90

and

96

? Use the

0.68-0.95-0.997

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

Calculate Standard Deviation: First, let's find out how many standard deviations away $90$ is from the mean. $\newline$ We do this by subtracting the mean from $90$ and dividing by the standard deviation: $(90 - 93) / 1 = -3$ .
Find Probability for $90$ : Now, let's do the same for $96$ . $\newline$ Subtract the mean from $96$ and divide by the standard deviation: $(96 - 93) / 1 = 3$ .
Find Probability for $96$ : So, we're looking for the probability that $X$ is between $-3$ and $3$ standard deviations from the mean. According to the $0.68-0.95-0.997$ rule, the probability that $X$ is within $3$ standard deviations (above or below) from the mean is about $0.997$ .
Calculate Probability Range: But we only need the probability from $-3$ to $3$ , not the full range from $-3$ to $+3$ . So we take half of $0.997$ , which is $0.997 / 2 = 0.4985$ .

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