The amount of fill(weight of contents)put into a glass jar of spaghetti sauce is normally distributed with mean=850g and standard deviation=8g.the probability that a random sample of 24 jars has a mean weight greater than 848g is?need help
Z = (sample mean-mean)/SEm (Standard Error of the mean)
SEm = SD/√(n-1)
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to that Z score.
To find the probability that a random sample of 24 jars has a mean weight greater than 848g, we can use the Central Limit Theorem. According to the Central Limit Theorem, if the sample size is large enough (typically n ≥ 30), the sample means will be approximately normally distributed regardless of the shape of the population distribution.
Since we are given that the population distribution of the fill weight is normally distributed with a mean of 850g and a standard deviation of 8g, we can consider the sample means to be normally distributed as well, with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.
Therefore, in this case, the sample means are normally distributed with a mean of 850g and a standard deviation of 8g / sqrt(24).
We need to find the probability that a random sample of 24 jars has a mean weight greater than 848g, which is equivalent to finding the area to the right of 848g on a normal distribution curve with a mean of 850g and a standard deviation of 8g / sqrt(24).
To find this probability, we can use a standard normal distribution table or a calculator that is capable of computing cumulative probabilities for a standard normal distribution.
Using the normal distribution table or calculator, we can find the z-score corresponding to the value of 848g. The z-score represents the number of standard deviations away from the mean.
The z-score can be calculated as follows:
z = (x - μ) / σ
where x is the observed value, μ is the population mean, and σ is the population standard deviation.
For our case, x = 848g, μ = 850g, and σ = 8g / sqrt(24).
Let's plug in the values and calculate the z-score:
z = (848 - 850) / (8 / sqrt(24))
z = -2 / (8 / sqrt(24))
z = -2 / (8 / 4.898979486)
z = -2 / 1.632993161
z ≈ -1.224744871
Now that we have the z-score, we can look up the corresponding area (probability) to the right of this z-score in the standard normal distribution table or use a calculator to find the cumulative probability.
Using a normal distribution table or calculator, we find that the probability associated with a z-score of -1.224744871 (or a mean weight greater than 848g) is approximately 0.109348.
Therefore, the probability that a random sample of 24 jars has a mean weight greater than 848g is approximately 0.109348 or 10.93%.