It looks like you've uploaded an image related to a probability distribution for the number of inquiries received by the Office of Admissions. You'd like to find the mean, variance, and standard deviation for this distribution. Let's go ahead and calculate those values.

Data from the image:

• Number of Inquiries (X): 22, 23, 24, 25, 26, 27
• Probability $P(X)$: 0.08, 0.19, 0.36, 0.25, 0.07, 0.05

We will use the following formulas:

• Mean (Expected Value): $\mu = \sum (X_i \cdot P(X_i))$

• Variance: $\sigma^2 = \sum \left[ P(X_i) \cdot (X_i - \mu)^2 \right]$

• Standard Deviation: $\sigma = \sqrt{\sigma^2}$

Let me calculate these values for you. numpy as np

Data

X = np.array([22, 23, 24, 25, 26, 27]) P_X = np.array([0.08, 0.19, 0.36, 0.25, 0.07, 0.05])

Mean (Expected Value)

mean = np.sum(X * P_X)

Variance

variance = np.sum(P_X * (X - mean)**2)

Standard Deviation

std_dev = np.sqrt(variance)

mean, variance, std_devThe calculations for the given probability distribution are as follows:

• Mean (Expected Value): $\mu = 24.19$
• Variance: $\sigma^2 = 1.45$
• Standard Deviation: $\sigma \approx 1.21$

Would you like further clarification on how these values were derived or have any other questions? Here are some related questions that might help:

1. How do you calculate the mean of a discrete probability distribution?
2. What is the relationship between variance and standard deviation in a probability distribution?
3. How would you calculate the mean and variance for a continuous probability distribution?
4. What is the importance of the standard deviation in statistics?
5. How do changes in probability values affect the mean and variance?

Tip: To understand how the standard deviation works, try imagining it as the "average distance" of each value from the mean. The higher the standard deviation, the more spread out the values are.