Respuesta :
Answer:
93.32% of the soda cans contain more than the advertised 12 ounces of soda
Step-by-step explanation:
Problems of normally distributed samples are 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:
[tex]\mu = 12.03, \sigma = 0.02[/tex]
What proportion of the soda cans contain more than the advertised 12 ounces of soda?
This is 1 subtracted by the pvalue off Z when X = 12. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{12 - 12.03}{0.02}[/tex]
[tex]Z = -1.5[/tex]
[tex]Z = -1.5[/tex] has a pvalue of 0.0668
1 - 0.0668 = 0.93332
93.32% of the soda cans contain more than the advertised 12 ounces of soda
Answer: the proportion of the soda cans that contain more than the advertised 12 ounces of soda is 0.933
Step-by-step explanation:
Since the amount of soda that the dispensing machine pours into a 12-ounce can of soda is normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = amount of soda poured into the soda cans
µ = mean amount
σ = standard deviation
From the information given,
µ = 12.03 ounces
σ = 0.02 ounce
The probability that the soda cans contain more than the advertised 12 ounces of soda is expressed as
P(x > 12) = 1 - P(x ≤ 12)
For x = 12,
z = (12 - 12.03)/0.02 = - 1.5
Looking at the normal distribution table, the probability corresponding to the z score is 0.067
P(x > 12) = 1 - 0.067
P(x > 12) = 0.933
