Question

The weights of a normally distributed group of adults participating in a fitness test has an average of 60kg and a standard deviation of 8kg.

a. What is the probability that an adult’s weight is in between 55-65kg?

b. What raw score divides the group such that 55% is above it?

Answer 1

Given, Average weight $\mu=60kg$

Standard deviation $\sigma=8kg$

(a) Probability that adult's weight lies in 55-65 kg =

$P(55<X<65)=P(\dfrac{55-60}{8}<z<\dfrac{65-60}{8})$

$=P(-0.625<z<0.625)$

$=1-2P(z>0.625)$

$=1-2(0.7324)=1-1.4648=0.5352$

(b) As the 55% groups above it

the z value at 0.55 is 0.7088

Then raw score

$X=\mu+z\times \sigma$

$=60+0.7088(8)=60+5.6704=65.6704.$

Hence The raw score is 65.6704.