The weight of a randomly selected green tea bag follows a normal distribution with mean of 53.0g and standard deviation of 2g.

1. 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.
2. 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 $(\mu )=53$

The population standard deviation $(\sigma )=2$

Solution:

A) The percentage of green tea bags with weights between 54g and 55g:
$$\therefore P(54<x<55)=P(\frac{54–53}{2}<\frac{x–\mu }{\sigma }<\frac{55–53}{2})$$ $$=P(0.5<z<1)$$ $$=P(z<1)–P(z<0.5)$$ $$=0.8413–0.6915$$ $$=0.1498$$ \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:

$$\therefore z=(x-\mu )/\sigma$$ $$\therefore 1.645=(x-53)/2$$ $$\therefore x=56.29$$