The new values for the mean and the standard deviation are 38 and 5 respectively.
The mean of a population is given as [tex]\mu = \frac{\sum_{N}^{i}x_i}{N}[/tex] and its standard deviation is given as [tex]\sigma = \sqrt{\frac{\sum_{N}^{i}(x_i-\mu)^2}{N}}[/tex].
In the question, we are given that a population has a mean of μ = 35 and a standard deviation of σ = 5, and are asked for the new values for the mean and standard deviation when 3 points are added to every score in the population.
We let the new population formed be over the variable k.
Thus, [tex]k_i = x_i + 3[/tex], for all i from 1 to N.
Thus, the new mean can be calculated as:
[tex]\mu_k = \frac{\sum_{N}^{i}k_i}{N}\\\mu_k = \frac{\sum_{N}^{i}(x_i + 3)}{N}\\\mu_k = \frac{\sum_{N}^{i}x_i + 3N}{N}\\\mu_k = \frac{\sum_{N}^{i}x_i}{N} + 3\\\mu_k = \mu + 3\\\mu_k = 35 + 3 = 38 [Since, \mu = 35].[/tex]
Thus, the new mean is 38.
Thus, the new standard deviation can be calculated as:
[tex]\sigma_k = \sqrt{\frac{\sum_{N}^{i}(k_i-\mu_k)^2}{N}}\\\sigma_k = \sqrt{\frac{\sum_{N}^{i}(x_i + 3-38)^2}{N}}\\\sigma_k = \sqrt{\frac{\sum_{N}^{i}(x_i-35)^2}{N}}\\\sigma_k = \sqrt{\frac{\sum_{N}^{i}(x_i-\mu)^2}{N}} [Since, \mu = 35]\\\sigma_k = \sigma = 5.[/tex]
Thus, the new standard deviation is 5.
Thus, the new values for the mean and the standard deviation are 38 and 5 respectively.
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