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find and interpret the standard deviation of the data set. 4. 18, 11, 1…

Question

find and interpret the standard deviation of the data set.

  1. 18, 11, 15, 20, 16
  2. 78, 71, 68, 75, 46, 66

Explanation:

Problem 4:

Step 1: Find the mean (\(\mu\))

The data set is \( 18, 11, 15, 20, 16 \). The number of data points \( n = 5 \).

$$ \mu=\frac{18 + 11+15 + 20+16}{5}=\frac{80}{5} = 16 $$

Step 2: Find the squared differences from the mean

For each data point \( x \), calculate \( (x-\mu)^2 \):

  • For \( x = 18 \): \( (18 - 16)^2=2^2 = 4 \)
  • For \( x = 11 \): \( (11 - 16)^2=(- 5)^2=25 \)
  • For \( x = 15 \): \( (15 - 16)^2=(-1)^2 = 1 \)
  • For \( x = 20 \): \( (20 - 16)^2=4^2=16 \)
  • For \( x = 16 \): \( (16 - 16)^2=0^2 = 0 \)

Step 3: Find the variance (\(\sigma^2\))

Variance is the average of the squared differences.

$$ \sigma^2=\frac{4 + 25+1+16 + 0}{5}=\frac{46}{5}=9.2 $$

Step 4: Find the standard deviation (\(\sigma\))

Standard deviation is the square root of the variance.

$$ \sigma=\sqrt{9.2}\approx3.24 $$
Problem 5:

Step 1: Find the mean (\(\mu\))

The data set is \( 78, 71, 68, 75, 46, 66 \). The number of data points \( n = 6 \).

$$ \mu=\frac{78+71 + 68+75+46+66}{6}=\frac{404}{6}\approx67.33 $$

Step 2: Find the squared differences from the mean

For each data point \( x \), calculate \( (x - \mu)^2 \):

  • For \( x = 78 \): \( (78-67.33)^2=(10.67)^2\approx113.85 \)
  • For \( x = 71 \): \( (71 - 67.33)^2=(3.67)^2\approx13.47 \)
  • For \( x = 68 \): \( (68 - 67.33)^2=(0.67)^2\approx0.45 \)
  • For \( x = 75 \): \( (75 - 67.33)^2=(7.67)^2\approx58.83 \)
  • For \( x = 46 \): \( (46 - 67.33)^2=(-21.33)^2\approx455.07 \)
  • For \( x = 66 \): \( (66 - 67.33)^2=(-1.33)^2\approx1.77 \)

Step 3: Find the variance (\(\sigma^2\))

$$ \sigma^2=\frac{113.85+13.47 + 0.45+58.83+455.07+1.77}{6}=\frac{643.44}{6} = 107.24 $$

Step 4: Find the standard deviation (\(\sigma\))

$$ \sigma=\sqrt{107.24}\approx9.72 $$

Answer:

The standard deviation is approximately \( 3.24 \). The data values are relatively close to the mean (average distance of about 3.24 from the mean).