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$X$ is a normally distributed random variable with mean $43$ and standard deviation $11$ . What is the probability that $X$ is between $41$ and $45$ ? 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

43

and standard deviation

11

. What is the probability that

X

is between

41

and

45

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

Given Data: Mean ( $\mu$ ) is $43$ and standard deviation ( $\sigma$ ) is $11$ . We need to find P(41 < X < 45).
Calculate z-score for $X = 41$ : First, calculate the z-score for $X = 41$ . $Z = \frac{X - \mu}{\sigma} = \frac{41 - 43}{11} = \frac{-2}{11} \approx -0.1818$ .
Calculate z-score for $X = 45$ : Now, calculate the z-score for $X = 45$ . $Z = \frac{X - \mu}{\sigma} = \frac{45 - 43}{11} = \frac{2}{11} \approx 0.1818$ .
Use $0.68$ - $0.95$ - $0.997$ Rule: Using the $0.68$ - $0.95$ - $0.997$ rule, we know that approximately $68\%$ of the data falls within one standard deviation of the mean. However, our $z$ -scores are less than $1$ , so we need to estimate the probability for this smaller range.
Estimate Probability Range: Since the z-scores $-0.1818$ and $0.1818$ are close to $0$ , we can estimate that the probability P(41 < X < 45) is less than $34\%$ (which is half of $68\%$ ). We need to look up the exact values in the z-table or use a calculator for more precision.
Find Exact Probabilities: After looking up the z-scores in the z-table or using a calculator, we find that the probability for $z = -0.1818$ is approximately $0.4281$ and for $z = 0.1818$ is approximately $0.5719$ .

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