Respuesta :
Answer:
Probability that a randomly selected can will have less than 15.5 ounces is 0.1587.
Step-by-step explanation:
We are given that the amount of soda in a 16-ounce can is normally distributed with a mean of 16 ounces and a standard deviation of 0.5 ounce.
Let X = amount of soda
So, X ~ N([tex]\mu=16,\sigma^{2} =0.5^{2}[/tex])
The z-score probability distribution for normal distribution is given by;
Z = [tex]\frac{ X -\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean amount = 16 ounces
[tex]\sigma[/tex] = standard deviation = 0.5 ounce
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.
So, the probability that a randomly selected can will have less than 15.5 ounces is given by = P(X < 15.5 ounces)
P(X < 15.5 ounces) = P( [tex]\frac{ X -\mu}{\sigma}[/tex] < [tex]\frac{ 15.5-16}{0.5}[/tex] ) = P(Z < -1) = 1 - P(Z [tex]\leq[/tex] 1)
= 1 - 0.8413 = 0.1587
Now, in the z table the P(Z [tex]\leq[/tex] x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1 in the z table which has an area of 0.8413.
Hence, the probability that a randomly selected can will have less than 15.5 ounces is 0.1587.
Using the normal distribution, it is found that there is a 0.1587 = 15.87% probability that a randomly selected can will have less than 15.5 ounces.
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:
- Mean of 16 ounces, thus [tex]\mu = 16[/tex].
- Standard deviation of 0.5 ounces, thus [tex]\sigma = 0.5[/tex].
The probability is the p-value of Z when X = 15.5, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{15.5 - 16}{0.5}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a p-value of 0.1587.
0.1587 = 15.87% probability that a randomly selected can will have less than 15.5 ounces.
A similar problem is given at https://brainly.com/question/24663213
