Question

Southwest Co-op produces bags of fertilizer, and it is concerned about impurity content. It is believed that the weights of impurities per bag are normally distributed with a mean of 12.2 grams and a standard deviation of 2.8 grams. A bag is chosen at random.
a. What is the probability that it contains less than 10 grams of impurities?
b. What is the probability that it contains more than 15 grams of impurities?
c. What is the probability that it contains between 12 and 15 grams of impurities?
d. It is possible, without doing the detailed calculations, to deduce which of the answers to parts (a) and (b) will be the larger. How would you do this?

Answer 1

Answer:

a) 0.216

b) 0.1587

c) 0.369

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 12.2 grams

Standard Deviation, σ = 2.8 grams

We are given that the distribution of weights of impurities per bag is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(contains less than 10 grams of impurities)

P(x < 10)

$P( x < 10) = P( z < \displaystyle\frac{10 - 12.2}{2.8}) = P(z < -0.7857)$

Calculation the value from standard normal z table, we have,  

$P(x < 10) =0.216= 21.6\%$

b) P(contains more than 15 grams of impurities)

P(x > 15)

$P( x > 15) = P( z > \displaystyle\frac{15 - 12.2}{2.8}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,  

$P(x > 610) = 1 - 0.8413 = 0.1587 = 15.87\%$

c) P(contains between 12 and 15 grams of impurities)

$P(12 \leq x \leq 15) = P(\displaystyle\frac{12 - 12.2}{2.8} \leq z \leq \displaystyle\frac{15-12.2}{2.8}) \\\\= P(-0.0714 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -0.0714)\\= 0.841 - 0.472 = 0.369 = 36.9\%$

$P(12 \leq x \leq 15) = 36.9%$

d) The mean divide the data in exactly two parts. Since 15 is farther away from the mean as compared to 10, the probability obtained in part (b) is smaller as compared to probability obtained in part (a).