Answer :
Answer:
The probability that the average weight of the boxes will exceed 94 lbs is 0.1587.
Step-by-step explanation:
Let X = weight of the boxes shipped.
The mean weight is, μ = 90 lbs and the standard deviation, σ = 24 lbs.
A sample of n = 36 boxes is selected.
According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and an appropriately huge random samples (n ≥ 30) is selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.
Then, the mean of the sample means is given by,
[tex]\mu_{\bar x}=\mu[/tex]
And the standard deviation of the sample means (also known as the standard error) is given by,
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]
Hence the sampling distribution of the sample mean weight of boxes is Normally distributed.
Compute the probability that the average weight of the boxes will exceed 94 lbs as follows:
[tex]P(\bar X>94)=P(\frac{\bar X-\mu_{\bar x}}{\sigma/\sqrt{n}}>\frac{94-90}{24/\sqrt{36}})\\=P(Z>1)\\=1-P(Z<1)\\=1-0.8413\\=0.1587[/tex]
Thus, the probability that the average weight of the boxes will exceed 94 lbs is 0.1587.
Final answer:
To find the probability that the average weight of the boxes will exceed 94 lbs, calculate the z-score and find the area under the normal curve. The probability is approximately 2.28%.
Explanation:
To find the probability that the average weight of the boxes will exceed 94 lbs, we need to calculate the z-score and then find the area under the normal curve. The z-score formula is given by: z = (x - μ) / (σ / sqrt(n)), where x is the desired value (94 lbs), μ is the population mean (90 lbs), σ is the population standard deviation (24 lbs), and n is the sample size (36 boxes).
Substituting the values into the formula, we get z = (94 - 90) / (24 / sqrt(36)) = 2. Therefore, we need to find the area to the right of z = 2 under the standard normal distribution.
Using a standard normal distribution table or calculator, we find that the probability of getting a z-score greater than 2 is approximately 0.0228 or 2.28%.
Learn more about Z-score calculation here:
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