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A set of five numbers is 8, 9, x, (x+1), 15. If the mean of these numbers is 11, find the value of x, the variance, and the standard deviation.

Mathematics

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Answers

2 parts
a

x=11x = 11

b

σ≈2.449\sigma \approx 2.449

Step-by-step solution

3 steps

Part a· 1 step

Step 1: Write the formula for the mean.

The mean of a set of numbers is the sum of the numbers divided by the count of the numbers.
The given numbers are 8,9,x,(x+1),158, 9, x, (x+1), 15. There are 5 numbers. The mean is 11.

Mean=∑xiN\text{Mean} = \frac{\sum x_i}{N} 11=8+9+x+(x+1)+15511 = \frac{8 + 9 + x + (x+1) + 15}{5}
  1. Solve for xx.

    Multiply both sides by 5:

    11×5=8+9+x+x+1+1511 \times 5 = 8 + 9 + x + x + 1 + 15 55=33+2x55 = 33 + 2x

    Subtract 33 from both sides:

    55−33=2x55 - 33 = 2x 22=2x22 = 2x

    Divide by 2:

    x=222x = \frac{22}{2} x=11\boxed{x = 11}

Part b

σ≈2.449\sigma \approx 2.449

Step 3: List the numbers with the value of xx.

Substitute x=11x=11 into the set of numbers:
8,9,11,(11+1),158, 9, 11, (11+1), 15
The numbers are 8,9,11,12,158, 9, 11, 12, 15.
The mean (μ\mu) is given as 11.

  1. Calculate the variance (σ2\sigma^2).

    The formula for variance is σ2=∑(xi−μ)2N\sigma^2 = \frac{\sum (x_i - \mu)^2}{N}.
    First, calculate the deviations from the mean (xi−μ)(x_i - \mu) and their squares (xi−μ)2(x_i - \mu)^2:

    • For 8: (8−11)2=(−3)2=9(8 - 11)^2 = (-3)^2 = 9
    • For 9: (9−11)2=(−2)2=4(9 - 11)^2 = (-2)^2 = 4
    • For 11: (11−11)2=(0)2=0(11 - 11)^2 = (0)^2 = 0
    • For 12: (12−11)2=(1)2=1(12 - 11)^2 = (1)^2 = 1
    • For 15: (15−11)2=(4)2=16(15 - 11)^2 = (4)^2 = 16

    Sum of the squared deviations:

    ∑(xi−μ)2=9+4+0+1+16=30\sum (x_i - \mu)^2 = 9 + 4 + 0 + 1 + 16 = 30

    Now, calculate the variance:

    σ2=305=6\sigma^2 = \frac{30}{5} = 6
  2. Calculate the standard deviation (σ\sigma).

    The standard deviation is the square root of the variance.

    σ=σ2\sigma = \sqrt{\sigma^2} σ=6\sigma = \sqrt{6} σ≈2.449\boxed{\sigma \approx 2.449}
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