Question

Assume the population has a distribution with a mean of 110 and a standard deviation of 8. For a random sample of size 40, find the 30th percentile of (round off to first decimal place).

Answer 1

Answer:

109.33

Step-by-step explanation:

given that the population has a distribution with a mean of 110 and a standard deviation of 8

Sample size n = 40

Std error of sample mean = std dev/square root of sample size

= $\frac{8}{\sqrt{40} } \\=1.265$

From std normal distribution table we find that 30th percentile for Z is

-0.5244

Corresponding x score =$110-0.5244*1.265=109.33663$

30th percentile = 109.33663

=109.33