(a) Approximately 68% of men have cranial capacities between ___ cc and ___ cc.
(b) Approximately ___ % of men have cranial capacities between 580 cc and 1780 cc.
Answer :
Given :
Population mean $(μ) = 1,180$
Population standard deviation $(σ) = 300$
Solution :
(a) Approximately 68% of men have cranial capacities between ___ cc and ___ cc.
$$\therefore \text{P}(-z < z < z) = 0.68$$ $$\therefore z = 0.994$$
$$\therefore \text{The middle 68%} = \mu \pm z \times \sigma$$ $$= 1,180 \pm 0.994 \times 300$$ $$= (881.8, 1,478.2)$$
B) The probability of men have cranial capacities between 580 cc and 1780 cc :
$$\therefore \text{P}(580 < x < 1,780) = \text{P}\left(\frac{580 – 1,180}{300} < \frac{x – \mu}{\sigma} < \frac{1,780 – 1,180}{300}\right)$$ $$= \text{P}(-2 < z < 2) = \text{P}(z < 2) – \text{P}(z < -2)$$ $$= 0.9772 – 0.0228$$ $$= 0.9544$$ $$\approx 95.44\%$$
