Question

The probability of women is distributed with the mean of 60kg and standard deviation 5kg. What is the probability that chosen at random

1. Greater than 70

2. Between 61&65

3. Less than 55

Answer 1

Solution:

Given, $\mu=60,\sigma=5$

$X\sim Bin(\mu,\sigma)$

1. $P(X>70)=P(z>\dfrac{70-60}{5})=P(z>2)$

$=1-P(z\le2)=1-0.97725=0.02275$

2. $P(61\le X\le 65)=P(\dfrac{61-60}{5}\le z\le \dfrac{65-60}{5})$

$=P(0.2\le z\le1)=P(z\le1)-P(z\le0.2)=0.84134-0.57926 \\=0.26208$

3. $P(X<55)=P(z<\dfrac{55-60}{5})=P(z<-1)=0.15866$