A Study Of Long-Term Phone Calls From General Electric

A Study Of Long Term Phone Calls Made From General Electricss Corpora

A study of long term phone calls made from General Electric's corporate headquarters in Fairfield, Connecticut, revealed that the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 4.2 minutes, and the standard deviation was 0.60 minutes.

The study aims to answer several questions related to the distribution of call durations:

- What fraction of calls last between 4.2 and 5 minutes?

- What fraction of calls last more than 5 minutes?

- What fraction of calls last between 5 and 6 minutes?

- What fraction of calls last between 4 and 6 minutes?

- What is the time corresponding to the longest 4% of calls?

Understanding these aspects provides insights into typical call durations and extreme cases, which is essential for resource planning, staff allocation, and optimizing communication strategies.

Paper For Above instruction

The analysis of call durations from General Electric's corporate headquarters employs principles of probability theory, specifically the properties of the normal distribution. Given that call lengths are normally distributed with a mean of 4.2 minutes and a standard deviation of 0.60 minutes, we can leverage the standard normal distribution (z-scores) to compute the desired fractions and cutoff points.

First, to compute the fraction of calls lasting between two times, we convert these times into z-scores using the formula:

\[ z = \frac{X - \mu}{\sigma} \]

where \(X\) is the call duration, \(\mu\) the mean, and \(\sigma\) the standard deviation. Using standard normal distribution tables or software, the cumulative probabilities associated with these z-scores give the fractions.

Question A: What fraction of calls last between 4.2 and 5 minutes?

Since 4.2 minutes equals the mean, its z-score is zero (\(z= (4.2-4.2)/0.6=0\)). For 5 minutes:

\[ z = \frac{5 - 4.2}{0.6} = \frac{0.8}{0.6} \approx 1.33 \]

Using standard normal tables or software, the cumulative probability for \(z=1.33\) is approximately 0.9082. The probability for \(z=0\) is 0.5, so:

\[ P(4.2

Thus, approximately 40.82% of calls last between 4.2 and 5 minutes.

Question B: What fraction of calls last more than 5 minutes?

Using the same z-score for 5 minutes (1.33), the cumulative probability is 0.9082; hence:

\[ P(X > 5) = 1 - P(Z

Approximately 9.18% of calls last more than 5 minutes.

Question C: What fraction of calls last between 5 and 6 minutes?

For 6 minutes:

\[ z = \frac{6 - 4.2}{0.6} = \frac{1.8}{0.6} = 3.0 \]

From standard normal tables, \(P(Z

\[ P(5

Approximately 9.05% of calls last between 5 and 6 minutes.

Question D: What fraction of calls last between 4 and 6 minutes?

For 4 minutes:

\[ z = \frac{4 - 4.2}{0.6} = -0.33 \]

From standard normal tables, \(P(Z

\[ P(4

Approximately 62.8% of calls last between 4 and 6 minutes.

Question E: What is the time corresponding to the longest 4% of calls?

The longest 4% are the upper 4% of the distribution. We find the z-score corresponding to the upper 4%, which is the 96th percentile:

\[ z_{0.96} \approx 1.75 \]

Using the z-score formula:

\[ X = \mu + z \times \sigma = 4.2 + 1.75 \times 0.6 = 4.2 + 1.05 = 5.25 \text{ minutes} \]

Thus, the longest 4% of calls last approximately 5.25 minutes or longer.

This analysis demonstrates the distribution of call durations, highlighting typical lengths and the extreme high-end calls, which could inform staffing and infrastructure planning.

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The second part of the analysis concerns the ATM machine near Kroger's in Union, Kentucky, with daily dispensed amounts modeled as normally distributed with a mean of $4,200 and a standard deviation of $720. The critical values are below $2,500 and above $6,000 for notifying the bank.

Question A: Percentage of days with dispensed amount less than $2,500.

Calculate z-score:

\[ z= \frac{2500 - 4200}{720} = \frac{-1700}{720} \approx -2.36 \]

From the standard normal table, \(P(Z

Question B: Percentage of days with dispensed amount exceeding $6,000.

Calculate z-score:

\[ z= \frac{6000 - 4200}{720} = \frac{1800}{720} = 2.50 \]

From tables, \(P(Z > 2.50) = 1 - P(Z

Approximately 0.62% of days, the bank is notified due to very high dispenses.

Question C: Percentage of days with dispensed amounts within the normal range (not triggering notifications).

Total percentage of days with normal amounts:

\[ 1 - (0.0091 + 0.0062) = 1 - 0.0153 = 0.9847 \]

or roughly 98.47%.

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The third segment involves the stock price of Bank of Florida over 240 trading days, modeled as normally distributed with a mean of $42.00 and a standard deviation of $2.25.

Question A: Percentage of days with stock price over $45.00.

Calculate z-score:

\[ z= \frac{45 - 42}{2.25} = 1.33 \]

From standard normal tables, \(P(Z > 1.33) = 1 - 0.9082 = 0.0918\), thus about 9.18% of days the stock price exceeds $45.00.

Number of days:

\[ 240 \times 0.0918 \approx 22 \]

Question B: Percentage of days with the stock price between $38 and $40.

Calculate z-scores:

For $38:

\[ z= \frac{38 - 42}{2.25} = -1.78 \]

For $40:

\[ z= \frac{40 - 42}{2.25} = -0.89 \]

From tables:

\(P(Z

\(P(Z

Difference:

\[ 0.1867 - 0.0375 = 0.1492 \]

Approximately 14.92% of days had prices between $38 and $40.

Question C: Highest 15% of stock prices.

Find the z-score corresponding to the 85th percentile:

\[ z_{0.85} \approx 1.04 \]

Calculate the stock price:

\[ X= \mu + z \times \sigma = 42 + 1.04 \times 2.25 \approx 42 + 2.34 = 44.34 \]

Therefore, on the top 15% of days, stock prices exceed approximately $44.34.

In conclusion, these statistical insights into call durations, transaction amounts, and stock prices enable better operational decision-making and risk assessment for corporations and financial institutions.

References

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