(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 $(μ) = 9$
The population standard deviation $(σ) = 0.9$
Solution :
a) The probability that x is over 10.3 kilograms :
$$\text{P}(x > 10.3) = \text{P} \left( \frac{x – \mu}{\sigma} > \frac{10.3 – 9}{0.9} \right)$$ $$= \text{P}(z > 1.444)$$ $$= 1 – \text{P}(z < 1.444)$$ $$= 1 – 0.9256$$ $$= 0.0744$$
b) The probability that x is at most 9.5 kilograms :
$$\text{P}(x \leq 9.5) = \text{P} \left( \frac{x – \mu}{\sigma} \leq \frac{9.5 – 9}{0.9} \right)$$ $$= \text{P}(z \leq 0.556)$$ $$= 0.7109$$
c) The probability that x is between 8.3 and 10.4 kilograms inclusive :
$$\text{P}(8.3 \leq x \leq 10.4) = \text{P} \left( \frac{8.3 – 9}{0.9} \leq \frac{x – \mu}{\sigma} \leq \frac{10.4 – 9}{0.9} \right)$$ $$= \text{P}(-0.778 \leq z \leq 1.556)$$ $$= \text{P}(z \leq 1.556) – \text{P}(z < -0.778)$$ $$= 0.9401 – 0.2183$$ $$= 0.7218$$