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Answer :
a) The sample standard deviation is 28.1 lbs.
b) 68% of the data lies within one standard deviation of the mean.
c) 95% of the data lies within two standard deviations of the mean.
(a) The mean and sample standard deviation of the weights:Mean:To find the mean, add up all the weights and then divide by the number of weights.μ = Σ xi / nμ = (179 + 172 + 153 + 197 + 175 + 83 + 169 + 162 + 168 + 189) / 10μ = 164.7
Therefore, the mean is 164.7 lbs.Standard Deviation:To calculate the sample standard deviation, we will use the formula: s = √((Σ (xi - μ)²) / (n - 1))
s = √((Σ (xi - μ)²) / (n - 1))
s = √(((179 - 164.7)² + (172 - 164.7)² + (153 - 164.7)² + (197 - 164.7)² + (175 - 164.7)² + (83 - 164.7)² + (169 - 164.7)² + (162 - 164.7)² + (168 - 164.7)² + (189 - 164.7)²) / (10 - 1))s = √((1263.61 + 92.41 + 126.09 + 870.49 + 119.69 + 3277.29 + 18.49 + 7.29 + 24.01 + 244.36) / 9)
s = √((6983.15) / 9)
s = 28.1
Therefore, the sample standard deviation is 28.1 lbs.
(b) What percent of the data lies within one standard deviation of the mean?For normally distributed data, we know that approximately 68% of the data lies within one standard deviation of the mean.
Therefore, approximately 68% of the data lies within one standard deviation of the mean.
(c) What percent of the data lies within two standard deviations of the mean?
For normally distributed data, we know that approximately 95% of the data lies within two standard deviations of the mean.
Therefore, approximately 95% of the data lies within two standard deviations of the mean.
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