Question

adimas found the mean of her 11 math test scores for the first semester.
overline { x } = \frac { ( 76 + 87 + 65 + 88 + 67 + 84 + 77 + 82 + 91 + 85 + 90 ) } { 11 } = \frac { 892 } { 11 } approx 81
using 81 as the mean, find the variance of her grades rounded to the nearest hundredth.
sigma ^ { 2 } =
find
ard deviation of her grades rounded to the nearest hundredth.
sigma =

Explanation:

Step1: Calculate the squared differences

For each score \(x_i\), calculate \((x_i - \bar{x})^2\).

• \( (76 - 81)^2=(- 5)^2 = 25\)
• \( (87 - 81)^2=(6)^2 = 36\)
• \( (65 - 81)^2=(-16)^2 = 256\)
• \( (88 - 81)^2=(7)^2 = 49\)
• \( (67 - 81)^2=(-14)^2 = 196\)
• \( (84 - 81)^2=(3)^2 = 9\)
• \( (77 - 81)^2=(-4)^2 = 16\)
• \( (82 - 81)^2=(1)^2 = 1\)
• \( (91 - 81)^2=(10)^2 = 100\)
• \( (85 - 81)^2=(4)^2 = 16\)
• \( (90 - 81)^2=(9)^2 = 81\)

Step2: Calculate the variance

The formula for variance \(\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n}\)
\(\sum_{i = 1}^{11}(x_i - \bar{x})^2=25+36 + 256+49+196+9+16+1+100+16+81=885\)
\(\sigma^{2}=\frac{885}{11}\approx80.45\)

Step3: Calculate the standard deviation

The formula for standard deviation \(\sigma=\sqrt{\sigma^{2}}\)
\(\sigma=\sqrt{80.45}\approx8.97\) (Note: There seems to be a mistake in the provided options as the correct variance is approximately \(80.45\) and standard deviation is approximately \(8.97\). If we assume the variance calculation is for a sample (but the problem uses \(\sigma^{2}\) which is population variance), if we follow the formula for population variance as above)

Answer:

Variance \(\sigma^{2}\approx80.45\), Standard deviation \(\sigma\approx8.97\)