QuIz Active TIME REMAININC 1 2 3 4 5 57:04 Grace is looking at a report of her monthly cell-phone usage for the last year to determine if she needs to upgrade her plan. The list represents the approximate number of megabytes of data Grace used each month. $$ \begin{array}{c} 700,735,680,890,755,740,670,785,805,1050,820, \ 750 \end{array} $$ What is the standard deviation of the data? Round to the nearest whole number. 65 75 100 130

Question

UpStudy AI · Accepted Answer

To determine the standard deviation of Grace's monthly cell-phone usage, follow these steps:

1. **List the Data:**
   $$
   700, 735, 680, 890, 755, 740, 670, 785, 805, 1050, 820, 750
   $$
   
2. **Calculate the Mean:**
   $$
   \text{Mean} = \frac{700 + 735 + 680 + 890 + 755 + 740 + 670 + 785 + 805 + 1050 + 820 + 750}{12} = \frac{9380}{12} \approx 781.67 \text{ MB}
   $$

3. **Compute Each Squared Deviation from the Mean:**
   $$
   \begin{align*}
   (700 - 781.67)^2 &\approx 6669.44 \
   (735 - 781.67)^2 &\approx 2177.78 \
   (680 - 781.67)^2 &\approx 10336.11 \
   (890 - 781.67)^2 &\approx 11736.11 \
   (755 - 781.67)^2 &\approx 711.11 \
   (740 - 781.67)^2 &\approx 1736.11 \
   (670 - 781.67)^2 &\approx 12369.44 \
   (785 - 781.67)^2 &\approx 11.11 \
   (805 - 781.67)^2 &\approx 544.44 \
   (1050 - 781.67)^2 &\approx 71957.44 \
   (820 - 781.67)^2 &\approx 1469.44 \
   (750 - 781.67)^2 &\approx 1002.78 \
   \end{align*}
   $$

4. **Sum of Squared Deviations:**
   $$
   \text{Total} \approx 120,721.32
   $$

5. **Calculate the Variance:**
   $$
   \text{Variance} = \frac{120,721.32}{12} \approx 10,060.11
   $$

6. **Determine the Standard Deviation:**
   $$
   \text{Standard Deviation} = \sqrt{10,060.11} \approx 100.3 \text{ MB}
   $$
   
   Rounding to the nearest whole number gives **100 MB**.

**Answer:** 100
The standard deviation of Grace's monthly cell-phone usage is **100 MB**.

Solución de tutoría real

Responder

Solución

Extra Insights

preguntas relacionadas

Latest Statistics Questions