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$X$ is a normally distributed random variable with mean $94$ and standard deviation $5$ . What is the probability that $X$ is between $89$ and $99$ ? 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

94

and standard deviation

5

. What is the probability that

X

is between

89

and

99

? Use the

0.68-0.95-0.997

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

Given Data: Mean ( $\mu$ ) is $94$ , standard deviation ( $\sigma$ ) is $5$ . We need to find P(89 < X < 99).
Calculate $k$ for $X=89$ : For $X = 89$ , calculate $k$ using $X = \mu + \sigma(k)$ . So, $89 = 94 + 5(k)$ .
Calculate $k$ for $X=99$ : Solving for $k$ gives us $k = \frac{89 - 94}{5}$ , which is $k = -1$ .
Calculate Probability: For $X = 99$ , calculate $k$ using $X = \mu + \sigma(k)$ . So, $99 = 94 + 5(k)$ .
Final Probability: Solving for $k$ gives us $k = (99 - 94) / 5$ , which is $k = 1$ .
Final Probability: Solving for $k$ gives us $k = (99 - 94) / 5$ , which is $k = 1$ .We have P(89 < X < 99) = P(\mu - \sigma < X < \mu + \sigma). According to the $0.68-0.95-0.997$ rule, this is approximately $0.68$ .
Final Probability: Solving for $k$ gives us $k = (99 - 94) / 5$ , which is $k = 1$ .We have P(89 < X < 99) = P(\mu - \sigma < X < \mu + \sigma). According to the $0.68-0.95-0.997$ rule, this is approximately $0.68$ .So, the probability that $X$ is between $89$ and $99$ is approximately $0.68$ .

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