Respuesta :
Answer:
12%
Step-by-step explanation:
Step 1
We find the standard deviation
A randomly selected 8-ounce can containing 35mg of caffeine is 1.2 standard deviations above the mean.
Mean = 33mg
35 = Mean - 1.2(standard deviation)
35 = 33 + 1.2(standard deviation)
35 - 33 = 1.2(standard deviation)
2 = 1.2(standard deviation)
Standard deviation = 2/1.2
= 1.6666666667
= 1.67
Therefore, the standard deviation is 1.67
Step 2
Using the z score formula
Z score = x - μ/σ
x = 35mg
μ = mean = 33mg
σ= standard deviation = 1.67
= 35 - 33/1.67
= 1.1976
Probabilty value from Z-Table:
P(x<35) = 0.88446
P(x>35) = 1 - P(x<35)
= 1 - 0.88446
= 0.11554
Converting to percentage = 0.11554 × 100
= 11.554%
≈ 12%
Therefore, the percent of 8-ounce cans of the cola having a caffeine content greater than 35mg is approximately 12%
Using the normal distribution, it is found that 11.5% of 8-ounce cans of the cola have a caffeine content greater than 35mg.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It 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, which is the percentile of X.
In this problem, a randomly selected 8-ounce can containing 35mg of caffeine is 1.2 standard deviations above the mean, which means that the z-score is Z = 1.2.
The percent of 8-ounce cans of the cola have a caffeine content greater than 35mg is 1 subtracted by the p-value of Z = 1.2, so:
Looking at the z-table, Z = 1.2 has a p-value of 0.885.
1 - 0.885 = 0.115
11.5% of 8-ounce cans of the cola have a caffeine content greater than 35mg.
A similar problem is given at https://brainly.com/question/24663213
