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In Exercises \(9-14, X\) has a normal distribution with the given mean and standard deviation. Find the indicated probabilities. $$ \mu=50, \sigma=10, \text { find } P(30 \leq X \leq 62) $$

Short Answer

Expert verified
The probability of X lying between 30 and 62 in this normal distribution is approximately \( P(30 \leq X \leq 62) = 0.8621 \) or 86.21%.
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Step 1: Identify the given information

The given information from the problem can be written as follows: \( \mu = 50 \), \( \sigma = 10 \) We are asked to find the probability \( P(30 \leq X \leq 62) \)

Step 2: Standardize the values using Z-score formula

The formula for calculating the Z-score is: \( Z = \frac{X - \mu}{\sigma} \) We need to find the Z-scores for both the lower bound (30) and upper bound (62). For the lower bound: \( Z_{lower} = \frac{30 - 50}{10} \) For the upper bound: \( Z_{upper} = \frac{62 - 50}{10} \)

Step 3: Calculate Z-scores

We now plug in the given values to calculate the Z-scores: Lower bound: \( Z_{lower} = \frac{30 - 50}{10} = -2 \) Upper bound: \( Z_{upper} = \frac{62 - 50}{10} = 1.2 \)

Step 4: Refer to the Z-table to find the respective probabilities

We can now look up the probabilities of the Z-scores in a Z-table: The probability that Z is less than or equal to -2 is approximately 0.0228 The probability that Z is less than or equal to 1.2 is approximately 0.8849

Step 5: Calculate the probability between the two Z-scores

Now, to find the probability \( P(30 \leq X \leq 62) \), we subtract the probability of Z being less than or equal to -2 from the probability of Z being less than or equal to 1.2: \( P(30 \leq X \leq 62) = P(Z \leq 1.2) - P(Z \leq -2) \) \( P(30 \leq X \leq 62) = 0.8849 - 0.0228 \)

Step 6: Final Answer

Our final answer is: \( P(30 \leq X \leq 62) = 0.8621 \) So the probability of X lying between 30 and 62 in this normal distribution is approximately 0.8621 or 86.21%.

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