asstignment 2

Question

a large group of public health university students took an examination in Biostatics and the final grades have a mean of 555 and a variance of 64. if we can approximate the distrbution of these grades by a normal distribution, what percent of the student:

1. scored higher than 60?

2. passed the examination (grades > 50)

3. failed the examination (grades < 50)

Answer 1

A variance (σ²) of 64 corresponds to a standard deviation (σ) of √64=8.

Mean μ=555. I'm guessing that 555 should be 55% or a score of 55.

(1) When X=60, Z=(60-μ)/σ for a large sample, that is, Z=(60-55)/8=⅝=0.625.

This Z-score corresponds to 0.734 approx or 73.40%. This means 73.40% of students at most score 60, implying that 26.6% have scores of 60 or more.

(2) and (3) When X=50, Z=(50-55)/8=-⅝=-0.625, corresponding to 26.6%. Therefore (2) 73.4% pass (above 50) while (3) 26.6% fail.