Answer to Question #223384 in Statistics and Probability for handsome

Question

A manufacturer fills jars with coffee. The weight of coffee W in grams in a jar can be modelled by a normal distribution with mean 232 grams standard deviation of 5 grams.

i) Find p(W is less than or equal to 224)

ii) Ten jars of coffee are selected at random what is the probability that their average content is between 228 and 235

Answer 1

i) Let $W=$ the weight of coffee: $W\sim N(\mu, \sigma^2).$

Given $\mu=232\ g, \sigma=5\ g.$

$$P(W\leq224)=P(Z\leq \dfrac{224-\mu}{\sigma})$$

$$=P(Z\leq \dfrac{224-232}{5})=P(Z\leq -1.6)$$

$$\approx0.0548$$

ii) Let $X=$ the average content of coffee: $X\sim N(\mu, \sigma^2/n).$

Given $\mu=232\ g, \sigma=5\ g, n=10$

$$P(228<X<235)=P(X<235)-P(X\leq 228)$$

$$=P(Z<\dfrac{235-232}{5/\sqrt{10}})-P(Z\leq\dfrac{228-232}{5/\sqrt{10}})$$

$$\approx P(Z<1.897367)-P(Z\leq -2.529822)$$

$$\approx0.971110-0.005706\approx0.9654$$