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Sagot :
Mean is 14.5 years and standard deviation is 0.34 years
The standard deviation is a measurement of how much a group of values vary or are dispersed. While a high standard deviation suggests that the values are dispersed throughout a wider range, a low standard deviation suggests that the values tend to be close to the established mean.
Given in the question:
Mean = 14.5 years, standard deviation, σ = 2.5 years and Sample size = 55 students.
The mean of the sample of 55 students will now stay the same for 14.5 years, and the formula will be used to determine the sample's standard deviation.:
= σ/√n
here n is the sample size
thus, Standard deviation of the sample of 55 students, σₓ = [tex]\frac{2.5}{\sqrt{55} }[/tex]
= 0.337 ≈ 0.34
Hence, the Mean = 14.5 years and standard deviation = 0.34 years.
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Mean of the sampling distribution is 14.5 . Standard Deviation of the sampling distribution is 0.45
The mean of the sampling distribution is equal to the mean of the population, which in this case is 14.5 years.
The standard deviation of the sampling distribution is calculated using the following formula:
Standard deviation of the sampling distribution [tex]= \frac{standard \ deviation \ of \ the \ population}{sample \ size}[/tex]
Plugging in the values given in the problem, we get:
Standard deviation of the sampling distribution = [tex]\frac{(2.5 years)} {\sqrt55}= 0.45[/tex] years
So the mean of the sampling distribution is 14.5 years, and the standard deviation of the sampling distribution is 0.45 years.
Standard deviation is a measure of the spread or dispersion of a set of data values. It is the square root of the variance, which is defined as the average of the squared differences from the mean. It is used to calculate how much variation exists from the average (mean) value of a set of data.
To learn more about mean, visit:
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