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?
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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.
Step 1: Input Parameters.
The provided dataset has the following characteristics:
- Mean (\( \mu \)) = 40,
- Variance (\( \sigma^2 \)) = 25,
- Standard deviation (\( \sigma \)) = \( \sqrt{25} = 5 \),
- Number of data points = 50.
The objective is to determine the probability that a randomly chosen data point falls within the range of 35 to 45.
Step 2: Z-Score Calculation.
To standardize the range [35, 45], we convert the boundary values into z-scores using the formula:
\[z = \frac{x - \mu}{\sigma}\]
where \( x \) represents a data value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
For the lower bound \( x = 35 \):
\[z_{35} = \frac{35 - 40}{5} = \frac{-5}{5} = -1\]
For the upper bound \( x = 45 \):
\[z_{45} = \frac{45 - 40}{5} = \frac{5}{5} = 1\]
Step 3: Probability Determination.
Using the standard normal distribution, we find the cumulative probabilities corresponding to the calculated z-scores:
- For \( z = -1 \), the cumulative probability \( P(z \leq -1) \) is approximately 0.1587.
- For \( z = 1 \), the cumulative probability \( P(z \leq 1) \) is approximately 0.8413.
The probability of a value lying between 35 and 45 is the difference between these cumulative probabilities:
\[P(35 \leq x \leq 45) = P(z \leq 1) - P(z \leq -1) = 0.8413 - 0.1587 = 0.6826\]
Answer: The probability that a randomly selected value from the dataset falls between 35 and 45 is approximately \( 0.68 \).