Answer:
58.
Step-by-step explanation:
We have been given that marks on a public health test follow a normal distribution with a mean of 77 and a standard deviation of 11. We are asked to find the approximate 40th percentile of the mark distribution.
We will use z-score formula and normal distribution table to solve our given problem.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
z = Z-score,
x = Sample score,
[tex]\mu[/tex] = Mean,
[tex]\sigma[/tex] = Standard deviation.
[tex]z=\frac{x-77}{11}[/tex]
From normal distribution table, we need to find z-score corresponding to 40th percentile or 0.40.
[tex]-1.75=\frac{x-77}{11}[/tex]
Let us solve for x.
[tex]-1.75*11=\frac{x-77}{11}*11[/tex]
[tex]-19.25=x-77[/tex]
[tex]-19.25+77=x-77+77[/tex]
[tex]57.75=x[/tex]
[tex]x\approx 58[/tex]
Therefore, the approximate 40th percentile of the mark distribution would be 58.
