Answer to Question #233006 in Statistics and Probability for Winsome Ogidi

Question

The weights of patients in a geriatric clinic are normally distributed with a mean of 60kg and a standard deviation of 10kg what is the probability that a patient picked at random has a weight between 50kg and 70kg

Answer 1

Let $X=$ weight of patient: $X\sim N(\mu, \sigma^2).$

Given $\mu=60\ kg, \sigma=10\ kg.$

$$P(50<X<70)=P(X<70)-P(X\leq50)$$

$$=P(Z<\dfrac{70-60}{10})-P(Z\leq\dfrac{50-60}{10})$$

$$=P(Z<1)-P(Z\leq-1)$$

$$\approx0.841344746-0.158655254$$

$$\approx0.6827$$

By 68–95–99.7 rule

$$P(50<X<70)=P(\mu-1\sigma<X<\mu+1\sigma)\approx$$

$$\approx0.6827$$