Respuesta :
The probability that randomly selected apple will contain more than [tex]2.50 \text{ }\text{ounce}[/tex] juice is [tex]\fbox{0.0475}[/tex].
Further explanation:
Given:
The mean [tex]\mu[/tex] of apple juice is [tex]2.50 \text{ }\text{ounce}[/tex].
The standard deviation [tex]\sigma[/tex] of apple juice is [tex]0.15 \text{ }\text{ounce}[/tex].
Calculation:
The random variable [tex]\bar{X}[/tex] follow Normal distribution with parameter mean and variance.
Normal distribution is symmetric and it bell shaped curve.
[tex]\bar{X} \sim \text{Normal}(\mu,\sigma^{2} )[/tex]
Variance is square of standard deviation.
Variance = [tex]\sigma^{2}[/tex]
Variance = [tex]0.15^{2}[/tex]
Variance = [tex]0.0225[/tex]
[tex]\bar{X}\sim \text{Normal}(\mu,\sigma^{2} )[/tex]
Substitute [tex]2.25[/tex] for [tex]\mu[/tex] and [tex]0.0225[/tex] for [tex]\sigma^{2}[/tex]
[tex]\bar{X}\sim\text {Normal}(2.25,0.0225)[/tex]
The probability that randomly selected apple will contain more than [tex]2.50[/tex] can be calculated by applying the central Limit Theorem as,
[tex]\text{Probability}=P(\bar{X}>2.50)}\\\text{Probability}=P\left(\dfrac{{\bar{X}-\mu}}{\sigma}>\dfrac{{2.50-2.25}}{0.15}\right)\\\text{Probability}=P\left(Z}>\dfrac{{0.25}}{0.15}\right)\\ \text{Probability}=P\left(Z}>\dfrac{{5}}{3}\right)\\\text{Probability}=P(Z}>1.67})[/tex]
The Normal distribution is symmetric.
Therefore, the probability of greater than [tex]1.67[/tex] is equal to the probability of less than [tex]1.67[/tex].
[tex]P(Z>1.67})=1-P(Z<1.67)\\P(Z>1.67})=1-0.9525\\P(Z>1.67})=0.0475[/tex]
Hence, the probability that randomly selected apple will contain more than [tex]2.50 \text{ }\text{ounce}[/tex] juice is [tex]\fbox{0.0475}[/tex].
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Answer Details:
Grade: College Statistics
Subject: Mathematics
Chapter: Probability and Statistics
Keywords:
Probability, Statistics, Apple juice, Normal distribution, Normal approximation, Central Limit Theorem, Z-table, Mean, Standard deviation, Symmetric, Variance, Conglomerate of Apple, Northwestern.
The distribution of apple follows a normal distribution.
The probability that a randomly selected apple will contain more than 2.50 ounces is 0.0475
The given parameters are:
[tex]\mu = 2.25[/tex] --- mean
[tex]\sigma = 0.15[/tex] --- standard deviation
[tex]x = 2.50[/tex]
First, we calculate the z score
[tex]z = \frac{x - \mu}{\sigma}[/tex]
Substitute known values
[tex]z = \frac{2.50 - 2.25}{0.15}[/tex]
[tex]z = \frac{0.25}{0.15}[/tex]
[tex]z = 1.67[/tex]
So, the required probability is:
[tex]P(X>x) = P(Z>z)[/tex]
This gives
[tex]P(x>2.50) = P(z>1.67)[/tex]
From z-score table, we have:
[tex]P(z>1.67) = 0.04746[/tex]
This means that:
[tex]P(x>2.50) =0.04746[/tex]
Approximate
[tex]P(x>2.50) =0.0475[/tex]
Hence, the probability that a randomly selected apple will contain more than 2.50 ounces is 0.0475
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