To find out how many data points Mona should expect to find between 1000 and 1800 in a normal distribution \( N(1400, 200) \), we first need to standardize the values using the z-score formula:
\[ z = \frac{(X - \mu)}{\sigma} \]
Where:
- \( X \) is the value we are interested in,
- \( \mu \) is the mean (1400),
- \( \sigma \) is the standard deviation (200).
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For \( X = 1000 \): \[ z = \frac{(1000 - 1400)}{200} = \frac{-400}{200} = -2 \]
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For \( X = 1800 \): \[ z = \frac{(1800 - 1400)}{200} = \frac{400}{200} = 2 \]
Next, we look up the z-scores in the standard normal distribution table (or use a calculator):
- The area to the left of \( z = -2 \) is approximately 0.0228 (or 2.28%).
- The area to the left of \( z = 2 \) is approximately 0.9772 (or 97.72%).
To find the area (or the proportion of data points) between z = -2 and z = 2: \[ P(-2 < Z < 2) = P(Z < 2) - P(Z < -2) = 0.9772 - 0.0228 = 0.9544 \]
Now, since Mona has 50 data points, we can calculate the expected number of data points between 1000 and 1800: \[ 0.9544 \times 50 \approx 47.22 \]
Rounding to the nearest whole number, we find that Mona should expect about 47 data points between 1000 and 1800.
So, the correct response is 47.
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