Apostmix beverage machine is adjusted to release a certain amount of syrup into a chamber where it is mixed with carbonated water. a random sample of 25 beverages was found to have a mean syrup content of x¯=1.19 fluid ounces and the sample standard deviation is s=0.015 fluid ounces. find a 95% two-sided confidence interval on the mean volume of syrup dispensed. assume population is approximately normally distributed.

  (1.1838, 1.1962)

Step-by-step explanation:

The formula we use to calculate the confidence interval for population mean ( if population standard deviation is not given)is given by :-

,

where n= sample size

s= sample standard deviation.

= sample mean

t* = Two-tailed critical t-value.

Given : n= 25

Degree of freedom : df = n-1 =24

Significance level

Now from students' t-distribution table , check the t-value for significance level and df=24:

t*=2.0639

 fluid ounces

s= 0.015  fluid ounces

We assume that the population is approximately normally distributed

Now, the 95% two-sided confidence interval on the mean volume of syrup dispensed :-

∴ The required confidence interval = (1.1838, 1.1962)

x+y = 29, 5x + 2y = 100

step-by-step explanation:

i'm guessing it was a multiple first option was the correct answer

boxplot e is the answer