SOLUTION: A large bank in South Africa knows that the average credit card debt owed by their card holders is R {1170} with a standard deviation of R {276}. In an investigation into the exten

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Question 1197726: A large bank in South Africa knows that the average credit card debt owed by their card holders is R {1170} with a standard deviation of R {276}. In an investigation into the extent of credit card debt at the bank, the bank collects random sample of {26} credit card accounts.
What is the probability that the mean credit card debt for the sample of accounts is more than R {992}?
What is the probability that the mean credit card debt for the sample of accounts is between R {1095} and R {1239}?

Answer by ElectricPavlov(122) About Me  (Show Source):
You can put this solution on YOUR website!
**1. Define Variables**
* **μ (mu):** Population mean credit card debt = R 1170
* **σ (sigma):** Population standard deviation of credit card debt = R 276
* **n:** Sample size = 26
* **x̄ (x-bar):** Sample mean credit card debt
**2. Calculate Standard Error of the Mean (SEM)**
* SEM = σ / √n
* SEM = 276 / √26
* SEM ≈ 54.08
**3. Standardize the Sample Mean (z-score)**
* **a) Probability of mean debt > R 992**
* z = (x̄ - μ) / SEM
* z = (992 - 1170) / 54.08
* z ≈ -3.29
* Using a z-table, P(z > -3.29) ≈ 0.9995
* **Therefore, the probability that the mean credit card debt for the sample is more than R 992 is approximately 0.9995 (or 99.95%).**
* **b) Probability of mean debt between R 1095 and R 1239**
* **i) For R 1095:**
* z = (1095 - 1170) / 54.08
* z ≈ -1.39
* **ii) For R 1239:**
* z = (1239 - 1170) / 54.08
* z ≈ 1.28
* Using a z-table:
* P(z < 1.28) ≈ 0.8997
* P(z < -1.39) ≈ 0.0823
* P(1095 < x̄ < 1239) = P(z < 1.28) - P(z < -1.39)
* P(1095 < x̄ < 1239) ≈ 0.8997 - 0.0823
* P(1095 < x̄ < 1239) ≈ 0.8174
* **Therefore, the probability that the mean credit card debt for the sample is between R 1095 and R 1239 is approximately 0.8174 (or 81.74%).**
**Key Assumptions:**
* The population of credit card debt amounts is normally distributed (or the sample size is large enough for the Central Limit Theorem to apply).
* The sample is a simple random sample.
**Note:**
* A z-table or statistical software can be used to find the probabilities associated with the z-scores.