Math Problem Statement

The table below gives the number of hours spent watching TV last week by a sample of 24 children.

55 49 60 85 73 72 67 55 44 32 81 69 27 86 44 70 90 60 83 63 60 29 92 19

Range =

Standard Deviation =

Round your answers to the nearest hundredth as needed.

Solution

Let's calculate the range and standard deviation for this dataset.

Step 1: Find the Range

The range is calculated by subtracting the minimum value from the maximum value.

  1. Identify the minimum value: 19
  2. Identify the maximum value: 92
  3. Calculate the range:
    Range=9219=73\text{Range} = 92 - 19 = 73

Step 2: Calculate the Standard Deviation

Step 2.1: Find the Mean

The mean (xˉ\bar{x}) is the average of all values.

xˉ=xn\bar{x} = \frac{\sum x}{n} where x\sum x is the sum of all values and nn is the number of values.

  1. Sum all values:
    55+49+60+85+73+72+67+55+44+32+81+69+27+86+44+70+90+60+83+63+60+29+92+19=130055 + 49 + 60 + 85 + 73 + 72 + 67 + 55 + 44 + 32 + 81 + 69 + 27 + 86 + 44 + 70 + 90 + 60 + 83 + 63 + 60 + 29 + 92 + 19 = 1300

  2. Divide by the number of values (24):
    xˉ=13002454.17\bar{x} = \frac{1300}{24} \approx 54.17

Step 2.2: Calculate Each Value's Deviation from the Mean, Square It, and Sum Them

(xxˉ)2\sum (x - \bar{x})^2 Calculate each deviation from the mean, square it, then sum these squared deviations.

Step 2.3: Divide by the Number of Values Minus 1 (Sample Standard Deviation)

The formula for the sample standard deviation ss is: s=(xxˉ)2n1s = \sqrt{\frac{\sum (x - \bar{x})^2}{n - 1}}

Let’s perform these calculations to find the standard deviation:

import numpy as np

Data set of hours spent watching TV

data = np.array([55, 49, 60, 85, 73, 72, 67, 55, 44, 32, 81, 69, 27, 86, 44, 70, 90, 60, 83, 63, 60, 29, 92, 19])

Calculate range

range_value = np.max(data) - np.min(data)

Calculate sample standard deviation (ddof=1 for sample)

standard_deviation = np.std(data, ddof=1)

range_value, round(standard_deviation, 2)(73, 20.77)(73, 20.77)successThe calculations yield the following results:

  • Range = 73
  • Standard Deviation ≈ 20.77

Would you like further details on these calculations or have any other questions?

Here are some related questions that might expand your understanding:

  1. How would the standard deviation change if the range were smaller?
  2. What is the impact of outliers on the standard deviation?
  3. Can the standard deviation be calculated for a population dataset differently?
  4. How do we interpret the range and standard deviation in terms of data spread?
  5. What other measures of dispersion are available besides range and standard deviation?

Tip: The standard deviation is a useful measure for understanding the spread of data around the mean, especially in contexts where we need to evaluate data consistency.

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Math Problem Analysis

Mathematical Concepts

Statistics
Range
Standard Deviation
Mean
Variance

Formulas

Range = max(x) - min(x)
Sample Standard Deviation: s = sqrt(sum((x - x̄)^2) / (n - 1))

Theorems

Standard deviation as a measure of spread
Range as a measure of spread

Suitable Grade Level

Grades 9-12