The weights of a large number of miniature poodles are approximately normally distributed with a mean of 9 kilograms and a standard deviation of 0.9 kilogram. If measurements are recorded to the nearest tenth of a kilogram, find the proportion of these poodles with weights:

(a) over 10.3 kilograms;

(b) of at most 9.5 kilograms;

(c) between 8.3 and 10.4 kilograms inclusive.

Answer :

Given :

The population mean $(\mu )=9$

The population standard deviation $(\sigma )=0.9$

Solution :

a) The probability that x is over 10.3 kilograms :

$$\text{P}(x>10.3)=\text{P}(\frac{x–\mu }{\sigma }>\frac{10.3–9}{0.9})$$ $$=\text{P}(z>1.444)$$ $$=1–\text{P}(z<1.444)$$ $$=1–0.9256$$ $$=0.0744$$

b) The probability that x is at most 9.5 kilograms :

$$\text{P}(x\le 9.5)=\text{P}(\frac{x–\mu }{\sigma }\le \frac{9.5–9}{0.9})$$ $$=\text{P}(z\le 0.556)$$ $$=0.7109$$

c) The probability that x is between 8.3 and 10.4 kilograms inclusive :

$$\text{P}(8.3\le x\le 10.4)=\text{P}(\frac{8.3–9}{0.9}\le \frac{x–\mu }{\sigma }\le \frac{10.4–9}{0.9})$$ $$=\text{P}(-0.778\le z\le 1.556)$$ $$=\text{P}(z\le 1.556)–\text{P}(z<-0.778)$$ $$=0.9401–0.2183$$ $$=0.7218$$