Question

In a dataset of 50 values, the mean is 40 and the variance is 25. What is the probability that a randomly selected value from this dataset is between 35 and 45?

• A. \( 0.68 \)
• B. \( 0.95 \)
• C. \( 0.34 \)
• D. \( 0.99 \)

Hint:

When working with normal distributions, use the z-score to standardize the values and find probabilities using the standard normal distribution table. For a normal distribution, approximately 68% of values lie within 1 standard deviation of the mean.

Answer 1

The Correct Option is A

Step 1: Given Information. The problem gives the following information: - Mean (\( \mu \)) = 40, - Variance (\( \sigma^2 \)) = 25, - Standard deviation (\( \sigma \)) = \( \sqrt{25} = 5 \), - The dataset consists of 50 values. We are asked to find the probability that a randomly selected value from this dataset lies between 35 and 45. Step 2: Convert the range to standard scores (z-scores). We can use the z-score formula to convert the values 35 and 45 into standard scores: \[ z = \frac{x - \mu}{\sigma} \] where: - \( x \) is the value from the dataset, - \( \mu \) is the mean, - \( \sigma \) is the standard deviation. For \( x = 35 \): \[ z_{35} = \frac{35 - 40}{5} = \frac{-5}{5} = -1 \] For \( x = 45 \): \[ z_{45} = \frac{45 - 40}{5} = \frac{5}{5} = 1 \] Step 3: Use the standard normal distribution. Now, we look up the z-scores in the standard normal distribution table: - For \( z = -1 \), the cumulative probability is approximately 0.1587. - For \( z = 1 \), the cumulative probability is approximately 0.8413. The probability that a value lies between 35 and 45 is the difference between these cumulative probabilities: \[ P(35 \leq x \leq 45) = P(z_{45}) - P(z_{35}) = 0.8413 - 0.1587 = 0.6826 \] Answer: Therefore, the probability that a randomly selected value from the dataset is between 35 and 45 is approximately \( 0.68 \).