- The percentage of green tea bags with weights between 54g and 55g is (a) 0.1587 (b) 0.3085 (c) 0.6915 (d) 0.8413 (e) 0.1498 13.
- The minimum weight to be one of the top 5% of heaviest green tea bags is (a) 56.92 (b) 49.08 (c) 56.29 (d) 49.71 (e) 89.185.
Answer:
Given Data :
The population mean $(μ)=53$
The population standard deviation $(\sigma) = 2$
Solution:
A) The percentage of green tea bags with weights between 54g and 55g:
$$\therefore P(54 < x < 55) = P\left(\frac{54 – 53}{2} < \frac{x – \mu}{\sigma} < \frac{55 – 53}{2}\right) $$ $$= P(0.5 < z < 1) $$ $$ = P(z < 1) – P(z < 0.5) $$ $$ = 0.8413 – 0.6915 $$ $$ = 0.1498$$ \approx 14.98% $$
B) The minimum weight to be one of the top 5% of heaviest green tea bags:
→ The z-score for right probability 0.05 = 1.645
Here we need to use a z -formula to find out the x value:
$$ \therefore z = (x-\mu)/\sigma $$ $$ \therefore 1.645 = (x-53)/2 $$ $$ \therefore x = 56.29 $$
