Answer:
[tex]\mu=36.57[/tex]
[tex]\sigma=11.43[/tex]
Step-by-step explanation:
We have been given that in a population of exam scores, a score of x=48 corresponds to z=+1.00 and a score of x=36 corresponds to z=-0.05.
To find the values of mean and standard deviation, we will use z-score formula as:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Let us solve for x (sample score).
[tex]x=z\cdot \sigma+\mu[/tex]
Upon substituting our given values, we will get two equations as:
[tex]48=1\cdot \sigma+\mu...(1)[/tex]
[tex]48-\mu=\sigma...(1)[/tex]
[tex]36=-0.05\cdot \sigma+\mu...(2)[/tex]
Upon substituting equation (1) in equation (2), we will get:
[tex]36=-0.05(48-\mu)+\mu[/tex]
[tex]36=-2.4+0.05\mu+\mu[/tex]
[tex]36+2.4=1.05\mu[/tex]
[tex]38.4=1.05\mu[/tex]
[tex]\frac{38.4}{1.05}=\frac{1.05\mu}{1.05}[/tex]
[tex]\mu=36.57142\approx 36.57[/tex]
Therefore, the mean of the given data would be approximately 36.57.
Upon substituting [tex]\mu=36.57[/tex] in equation (1), we will get:
[tex]48-36.57=\sigma[/tex]
[tex]\sigma=11.43[/tex]
Therefore, the standard deviation for the given data would be approximately 11.43.
