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Problem 11

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

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%.
See the step by step solution

## 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: $$P(30 \leq X \leq 62) = P(Z \leq 1.2) - P(Z \leq -2)$$ $$P(30 \leq X \leq 62) = 0.8849 - 0.0228$$

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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