Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = 16, \sigma = 0.31[/tex].
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 [tex]X = 16.62[/tex]. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{16.62 - 16}{0.31}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] 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 [tex]\mu[/tex], with [tex]X = 16.62[/tex] when Z has a pvalue of 0.94. It is between [tex]Z = 1.55[/tex] and [tex]Z = 1.56[/tex], so we use [tex]Z = 1.555[/tex].
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.555 = \frac{16.62 - \mu}{0.31}[/tex]
[tex]\mu = 16.62 - 0.31*1.555[/tex]
[tex]\mu = 16.14[/tex]
The mean amount dispensed by the machine should be set at 16.14 ounces to satisfy this wish.
By using z-score formula we got that The probability that a new cup will overflow when filled by the automatic dispenser is 0.0228 and If probability that a new cup will overflow is .006 then value from which should the mean amount dispensed by the machine be set to satisfy this wish is 16.14
What is z-score formula?
The Z-score measures how many standard deviations the measure is from the mean
[tex]Z=\frac{X-\mu}{\sigma}[/tex]
Given that
[tex]$\mu=16, \sigma=0.31$[/tex]
The probability that a new cup will overflow when filled by the automatic dispenser can be calculated as
[tex]$\begin{aligned}&Z=\frac{X-\mu}{\sigma} \\&Z=\frac{16.62-16}{0.31} \\&Z=2\end{aligned}$[/tex]
So by normal distribution table
[tex]P=$1-0.9772=0.0228$[/tex]
If probability that a new cup will overflow is .006 then value from which should the mean amount dispensed by the machine be set to satisfy this wish can be calculated as
[tex]$\begin{aligned}&Z=\frac{X-\mu}{\sigma} \\&1.555=\frac{16.62-\mu}{0.31} \\&\mu=16.62-0.31 * 1.555 \\&\mu=16.14\end{aligned}$[/tex]
By using z-score formula we got that The probability that a new cup will overflow when filled by the automatic dispenser is 0.0228 and If probability that a new cup will overflow is .006 then value from which should the mean amount dispensed by the machine be set to satisfy this wish is 16.14
To learn more about Z-score visit: https://brainly.com/question/25638875