Question

The number of potato chips in a bag is normally distributed with a mean of 74 and and a standard deviation of 4. Approximately what percent of bags contain between 62 and86 potato chips?

Answer 1

Answer:

99.8%  is the percent of bags contain between 62 and 86 potato chips.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 74

Standard Deviation, σ = 4

We are given that the distribution of number of potato chips in a bag is a bell shaped distribution that is a normal distribution.

Formula:

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

P( bags contain between 62 and 86 potato chips)

$P(62 \leq x \leq 86) = P(\displaystyle\frac{62 - 74}{4} \leq z \leq \displaystyle\frac{86-74}{4}) = P(-3 \leq z \leq 3)\\\\= P(z \leq 3) - P(z < -3)\\= 0.999 - 0.001 = 0.998 = 99.8\%$

$P(62 \leq x \leq 86) = 99.8\%$

Answer 2

Answer:

99.7%  is the percent of bags contain between 62 and 86 potato chips.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 74

Standard Deviation, σ = 4

We are given that the distribution of number of potato chips in a bag is a bell shaped distribution that is a normal distribution.

Formula:

P( bags contain between 62 and 86 potato chips)