Question

The amount of corn chips dispensed into a bag by the dispensing machine has been identified as possessing a normal distribution with a mean of μ=48.5 ounces and a standard deviation of σ=0.2 ounce. What chip amount represents the 67th percentile, p 67, for the bag weight distribution? Round to the nearest hundredth. Hint: the 67th percentile of the standard normal curve is z=0.44. Round your answer to to decimal places.

Answer 1

Answer:

48.588 ounce

Step-by-step explanation:

The amount of corn chips dispensed into a bag by the dispensing machine has been identified as possessing a normal distribution with a mean of μ=48.5 ounces and a standard deviation of σ=0.2 ounce.

We want to find the chip amount that represents the 67th percentile.

We use the formula:

$z = \frac{x - \mu}{ \sigma}$

We substitute μ=48.5, σ=0.2 and z=0.44 to obtain:

$0.44 = \frac{x - 48.5}{0.2}$

$0.2 \times 0.44 = x - 48.5$

$0.088 = x - 48.5$

$x = 48.5 + 0.088$

$x = 48.588$

The chip amount that represents the amount of the 67th percentile is 48.588 ounce

Answer 2

The chip amount that represents the 67th percentile is 48.588.and this can be determined by using the formula of z-score.

Given :

The amount of corn chips dispensed into a bag by the dispensing machine has been identified as possessing a normal distribution with a mean of μ = 48.5 ounces and a standard deviation of σ = 0.2 ounces.

To determine the chip amount that represents the 67th percentile, the below formula can be used:

$\rm z = \dfrac{x-\mu}{\sigma}$

Now, substitute the values of known terms in the above formula:

$\rm 0.44 = \dfrac{x - 48.5}{0.2}$

Cross multiply in the above equation.

$\rm 0.44\times 0.2 = x - 48.5$

Now further, simplify the above equation.

0.088 = x - 48.5

x = 48.5 + 0.088

x = 48.588

So, the chip amount that represents the 67th percentile is 48.588.

For more information, refer to the link given below:

https://brainly.com/question/23017717