Am i right?? please help!

Someone help me and remember to show your work so I know how to do it!

Monica [59]

(x^15)(x^-3) = x^(15 + (-3)) = x^12
n = 12

6 0

4 years ago

The Nelson Company makes the machines that automatically dispense soft drinks into cups. Many national fast food chains such as

Vika [28.1K]

Answer:

a) There is a 2.28% probability that a new cup will overflow when filled by the automatic dispenser.

b) The mean amount dispensed by the machine should be set at 16.14 ounces to satisfy this wish.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

Normal model with mean 16 ounces and standard deviation 0.31 ounces. This means that $\mu = 16, \sigma = 0.31$ .

A new 16-ounce cup that is being considered for use actually holds 16.62 ounces of drink.

a. What is the probability that a new cup will overflow when filled by the automatic dispenser?

This probability is 1 subtracted by the pvalue of Z when $X = 16.62$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{16.62 - 16}{0.31}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772. This means that there is a 1-0.9772 = 0.0228 = 2.28% probability that a new cup will overflow when filled by the automatic dispenser.

b. The company wishes to adjust the dispenser so that the probability that a new cup will overflow is .006. At what value should the mean amount dispensed by the machine be set to satisfy this wish?

This is the value of $\mu$ , with $X = 16.62$ when Z has a pvalue of 0.94. It is between $Z = 1.55$ and $Z = 1.56$ , so we use $Z = 1.555$ .