Question

(3) A survey revealed that the mean cholesterol level of male athletes in Ghana aged 20 – 30 years is 180 mg/dl and the standard deviation is 43 mg/dl. If a random sample of 60 male athletes is selected from this population, what is the probability that the sample mean cholesterol level will be greater than 190 mg/dl?

Answer 1

given : mean = 180

Sd= 43, n= 60

to find :

P($\overline{X}$ > 190)

solution :

P($\overline{X}$ > 190)= P($\frac{\overline{X}-mean}{\frac{sd}{\sqrt{n}}}$ > $\frac{190-180}{\frac{43}{\sqrt{60}}}$ )

=P(Z>1.80)

=1- P(Z<1.80)

Using z table , 1- 0.9642

= 0.0358