Questions: The relative frequency histogram represents the length of phone calls on George's cell phone during the month of September. Determine whether or not the histogram indicates that a normal distribution could be used as a model for the variable.

The relative frequency histogram represents the length of phone calls on George's cell phone during the month of September. Determine whether or not the histogram indicates that a normal distribution could be used as a model for the variable.

Transcript text: The relative frequency histogram represents the length of phone calls on George's cell phone during the month of September. Determine whether or not the histogram indicates that a normal distribution could be used as a model for the variable.

Solution

Solution Steps

Step 1: Calculate the Mean

The mean (\( \mu \)) of the phone call lengths is calculated as follows:

\[ \mu = \frac{\sum_{i=1}^N x_i}{N} = \frac{100}{5} = 20.0 \]

Step 2: Calculate the Variance and Standard Deviation

The variance (\( \sigma^2 \)) is calculated using the formula:

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} = 62.5 \]

The standard deviation (\( \sigma \)) is then:

\[ \sigma = \sqrt{62.5} \approx 7.91 \]

Step 3: Perform the Chi-Square Goodness-of-Fit Test

The Chi-Square test statistic (\( \chi^2 \)) is calculated as:

\[ \chi^2 = \sum_i \frac{(O_i - E_i)^2}{E_i} = 0.0 \]

The critical value for a Chi-Square test with 4 degrees of freedom at \( \alpha = 0.05 \) is:

\[ \chi^2(0.95, 4) = 9.4877 \]

Step 4: Calculate the P-Value

The P-Value associated with the Chi-Square statistic is:

\[ P = P(\chi^2 > 0.0) = 1.0 \]

Step 5: Conclusion

Since the P-Value \( (1.0) \) is greater than \( \alpha = 0.05 \), we fail to reject the null hypothesis. This indicates that the histogram suggests a normal distribution could be used as a model for the variable.