Question 1193206
**a. Estimate of the proportion of employees who do not use the ATM in a month**

* **Calculate the number of employees who do not use the ATM:** 
    * Number of employees who do not use the ATM = 20 employees (from the table)

* **Estimate the proportion:**
    * Proportion = (Number of employees who do not use the ATM) / (Total number of employees in the sample) 
    * Proportion = 20 / 100 = 0.20

**Therefore, the estimate of the proportion of employees who do not use the ATM in a month is 0.20.**

**b-1. 90% Confidence Interval for the Proportion**

* **Find the sample proportion (p̂):**
    * p̂ = 0.20 

* **Find the standard error (SE):**
    * SE = √[(p̂ * (1 - p̂)) / n] 
    * SE = √[(0.20 * 0.80) / 100] 
    * SE = √(0.16 / 100) 
    * SE = √0.0016 
    * SE = 0.04

* **Find the critical value (zα/2) for a 90% confidence level:**
    * For a 90% confidence level, α = 0.10 
    * zα/2 = z0.05 
    * Using a standard normal distribution table, we find z0.05 ≈ 1.645

* **Calculate the margin of error (E):**
    * E = zα/2 * SE 
    * E = 1.645 * 0.04 
    * E ≈ 0.0658

* **Calculate the confidence interval:**
    * Lower limit = p̂ - E = 0.20 - 0.0658 = 0.1342 
    * Upper limit = p̂ + E = 0.20 + 0.0658 = 0.2658

**The 90% confidence interval for the proportion of employees who do not use the ATM in a month is (0.134, 0.266).**

**b-2. Can National be sure that at least 40% of the employees of Fun Toy Company will use the ATM?**

* No, National cannot be sure that at least 40% of the employees will use the ATM. 
* The lower limit of the 90% confidence interval for the proportion of employees who do not use the ATM is 0.134. 
* This means that there is a possibility that the true proportion of employees who do not use the ATM could be as high as 13.4%. 
* If 13.4% of employees do not use the ATM, then only 86.6% would use it, which is below 90%.

**c. How many transactions does the average Fun employee make per month?**

* **Calculate the mean number of transactions:**

    * Mean = Σ(x * f(x)) / Σf(x) 

        * Where:
            * x represents the number of ATM uses 
            * f(x) represents the frequency of each number of uses

    * Mean = (0 * 20 + 1 * 30 + 2 * 20 + 3 * 5 + 4 * 5 + 5 * 20) / 100
    * Mean = (0 + 30 + 40 + 15 + 20 + 100) / 100
    * Mean = 205 / 100 
    * Mean = 2.05

**The average Fun employee makes 2.05 transactions per month.**

**d. 90% Confidence Interval for the Mean Number of Transactions**

* **Calculate the sample standard deviation (s):** 

    * This requires calculating the variance first: 
        * Variance (s²) = Σ[(x - Mean)² * f(x)] / (n - 1) 
        * Variance (s²) = [(0 - 2.05)² * 20 + (1 - 2.05)² * 30 + ... + (5 - 2.05)² * 20] / (100 - 1) 
        * Variance (s²) ≈ 2.3025 

    * Standard deviation (s) = √s² 
    * Standard deviation (s) = √2.3025 
    * Standard deviation (s) ≈ 1.517

* **Calculate the standard error of the mean (SEM):**
    * SEM = s / √n 
    * SEM = 1.517 / √100 
    * SEM = 1.517 / 10 
    * SEM = 0.1517

* **Find the critical value (tα/2) for a 90% confidence level with 99 degrees of freedom (n - 1):**
    * Using a t-distribution table, we find tα/2 ≈ 1.660 

* **Calculate the margin of error (E):**
    * E = tα/2 * SEM 
    * E = 1.660 * 0.1517 
    * E ≈ 0.2518

* **Calculate the confidence interval:**
    * Lower limit = Mean - E = 2.05 - 0.2518 = 1.798
    * Upper limit = Mean + E = 2.05 + 0.2518 = 2.302

**The 90% confidence interval for the mean number of transactions per month is (1.798, 2.302).**