Question 1167102: The manager of the commercial mortgage department of a large bank has collected data during the past two years concerning the number of commercial mortgages approved per week. The results from these two years (104 weeks) are shown to the right.
a. Compute the expected number of mortgages approved per week.
b. Compute the standard deviation.
c. What is the probability that there will be more than one commercial mortgage approved in a given week?
Number Approved , Frequency
0, 13
1, 25
2,33
3, 16
4, 8
5, 6
6 ,2
7, 1
a. The expected number of mortgages approved per week is
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! ```python?code_reference&code_event_index=2
import pandas as pd
import numpy as np
# Data
data = {
'X': np.arange(0, 8), # Number Approved (0, 1, 2, 3, 4, 5, 6, 7)
'Frequency': np.array([13, 25, 33, 16, 8, 6, 2, 1])
}
df = pd.DataFrame(data)
# Total number of weeks
N = df['Frequency'].sum()
# 1. Calculate the Probability P(X)
df['P(X)'] = df['Frequency'] / N
# --- a. Compute the Expected Number of Mortgages (Mean, E[X]) ---
df['X*P(X)'] = df['X'] * df['P(X)']
E_X = df['X*P(X)'].sum()
# --- b. Compute the Standard Deviation (sigma) ---
# E[X^2] is required for variance
df['X^2'] = df['X']**2
df['X^2*P(X)'] = df['X^2'] * df['P(X)']
E_X2 = df['X^2*P(X)'].sum()
# Variance: Var(X) = E[X^2] - (E[X])^2
variance = E_X2 - (E_X)**2
# Standard Deviation: sigma = sqrt(Var(X))
sigma = np.sqrt(variance)
# --- c. What is the probability that there will be more than one commercial mortgage approved in a given week? ---
# P(X > 1) = 1 - P(X <= 1) = 1 - P(X=0) - P(X=1)
P_X_leq_1 = df.loc[df['X'] <= 1, 'P(X)'].sum()
P_X_gt_1 = 1 - P_X_leq_1
# Print results
print(f"Total Weeks (N): {N}")
print(f"Expected Value (E[X]): {E_X:.2f}")
print(f"Standard Deviation (sigma): {sigma:.2f}")
print(f"P(X > 1): {P_X_gt_1:.4f}")
print("\nDataFrame with intermediate calculations:")
print(df.to_string())
```
```text?code_stdout&code_event_index=2
Total Weeks (N): 104
Expected Value (E[X]): 2.12
Standard Deviation (sigma): 1.50
P(X > 1): 0.6346
DataFrame with intermediate calculations:
X Frequency P(X) X*P(X) X^2 X^2*P(X)
0 0 13 0.125000 0.000000 0 0.000000
1 1 25 0.240385 0.240385 1 0.240385
2 2 33 0.317308 0.634615 4 1.269231
3 3 16 0.153846 0.461538 9 1.384615
4 4 8 0.076923 0.307692 16 1.230769
5 5 6 0.057692 0.288462 25 1.442308
6 6 2 0.019231 0.115385 36 0.692308
7 7 1 0.009615 0.067308 49 0.471154
```
Based on the provided data from $104$ weeks, here are the required statistical measures:
## a. The expected number of mortgages approved per week
The **expected number** ($E[X]$, or mean) is the weighted average of the number of approvals, where the weights are the probabilities (frequencies divided by 104).
$$E[X] = \sum x \cdot P(x)$$
$$E[X] \approx \mathbf{2.12}$$
## b. Compute the standard deviation
The **standard deviation** ($\sigma$) measures the spread or variability in the number of approvals around the mean.
$$\sigma = \sqrt{\sum (x - E[X])^2 \cdot P(x)}$$
$$\sigma \approx \mathbf{1.50}$$
## c. What is the probability that there will be more than one commercial mortgage approved in a given week?
This is the probability $P(X > 1)$, which is the sum of the probabilities for 2, 3, 4, 5, 6, or 7 approvals.
It's easier to calculate this as $1 - P(X \le 1)$:
$$P(X > 1) = 1 - [P(X=0) + P(X=1)]$$
$$P(X=0) = \frac{13}{104} \approx 0.1250$$
$$P(X=1) = \frac{25}{104} \approx 0.2404$$
$$P(X > 1) = 1 - (0.1250 + 0.2404)$$
$$P(X > 1) = 1 - 0.3654$$
$$P(X > 1) \approx \mathbf{0.6346}$$
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