adimas found the mean of her 11 math test scores for the first semester.\n overline { x } = \frac { ( 76 +…
adimas found the mean of her 11 math test scores for the first semester.\n overline { x } = \frac { ( 76 + 87 + 65 + 88 + 67 + 84 + 77 + 82 + 91 + 85 + 90 ) } { 11 } = \frac { 892 } { 11 } approx 81 \nusing 81 as the mean, find the variance of her grades rounded to the nearest hundredth.\n sigma ^ { 2 } = \nfind the standard deviation of her grades rounded to the nearest hundredth.\n sigma =
Answer
Explanation:
Step1: Calculate ((x_i - \bar{x})^2) for each (x_i)
- For (x_1 = 76): ((76 - 81)^2=(- 5)^2 = 25)
- For (x_2 = 87): ((87 - 81)^2=(6)^2 = 36)
- For (x_3 = 65): ((65 - 81)^2=(-16)^2 = 256)
- For (x_4 = 88): ((88 - 81)^2=(7)^2 = 49)
- For (x_5 = 67): ((67 - 81)^2=(-14)^2 = 196)
- For (x_6 = 84): ((84 - 81)^2=(3)^2 = 9)
- For (x_7 = 77): ((77 - 81)^2=(-4)^2 = 16)
- For (x_8 = 82): ((82 - 81)^2=(1)^2 = 1)
- For (x_9 = 91): ((91 - 81)^2=(10)^2 = 100)
- For (x_{10}=85): ((85 - 81)^2=(4)^2 = 16)
- For (x_{11}=90): ((90 - 81)^2=(9)^2 = 81)
Step2: Calculate the variance (\sigma^{2})
The formula for variance (\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n}) (\sum_{i = 1}^{11}(x_i - 81)^2=25+36 + 256+49+196+9+16+1+100+16+81=885) (\sigma^{2}=\frac{885}{11}\approx80.45)
Step3: Calculate the standard deviation (\sigma)
The formula for standard deviation (\sigma=\sqrt{\sigma^{2}}) (\sigma=\sqrt{\frac{885}{11}}\approx8.97)
Answer:
The variance (\sigma^{2}\approx80.45) and the standard deviation (\sigma\approx8.97)