Coal is carried from a mine in West Virginia to a power plant in New York in hopper cars on a long train. The automatic hopper car loader is set to put 55 tons of coal into each car. The actual weights of coal loaded into each car are normally distributed, with mean ì = 55 tons and standard deviation ó = 0.8 ton.
What is the probability that one car chosen at random will have less than 54.5 tons of coal? (Round your answer to four decimal places.)
What is the probability that 37 cars chosen at random will have a mean load weight x bar of less than 54.5 tons of coal? (Round your answer to four decimal places.)
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1. Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
2. Z = (score-mean)/SEm
SEm = SD/√n
Use same table.
To find the probability that one car chosen at random will have less than 54.5 tons of coal, we can use the standard normal distribution.
Step 1: Calculate the z-score.
The z-score formula is:
z = (x - μ) / σ
where x is the value we are interested in, μ is the mean, and σ is the standard deviation.
In this case, x = 54.5 tons, μ = 55 tons, and σ = 0.8 ton.
Using the formula, we can calculate the z-score as:
z = (54.5 - 55) / 0.8
Step 2: Look up the z-score in the standard normal distribution table.
The standard normal distribution table gives the area to the left of a given z-score.
Looking up the z-score of -0.625 (rounded to three decimal places) in the table, we find the corresponding area is 0.2659.
Step 3: Calculate the probability.
Since we are looking for the probability that the car has less than 54.5 tons of coal, we need to find the area to the left of the z-score.
Therefore, the probability that one car chosen at random will have less than 54.5 tons of coal is approximately 0.2659.
To find the probability that 37 cars chosen at random will have a mean load weight less than 54.5 tons of coal, we can use the central limit theorem.
Step 1: Calculate the standard error of the mean.
The standard error of the mean (SE) is calculated as:
SE = σ / sqrt(n)
where σ is the standard deviation (0.8 ton) and n is the sample size (37 cars).
SE = 0.8 / sqrt(37)
Step 2: Calculate the z-score.
The z-score formula is the same as before:
z = (x - μ) / SE
In this case, x = 54.5 tons, μ = 55 tons, and SE = 0.8 / sqrt(37).
Using the formula, we can calculate the z-score as:
z = (54.5 - 55) / (0.8 / sqrt(37))
Step 3: Look up the z-score in the standard normal distribution table.
Looking up the z-score in the table, we find the corresponding area to the left of the z-score.
Step 4: Calculate the probability.
Since we are looking for the probability that the mean load weight is less than 54.5 tons of coal, we need to find the area to the left of the z-score.
Therefore, the probability that 37 cars chosen at random will have a mean load weight x-bar of less than 54.5 tons of coal is the area to the left of the z-score obtained in Step 3.
To find the probability that one car chosen at random will have less than 54.5 tons of coal, we need to calculate the Z-score and then use the standard normal distribution table.
1. First, calculate the Z-score using the formula:
Z = (X - μ) / σ
where X is the value we want to find the probability for (54.5 tons), μ is the mean (55 tons), and σ is the standard deviation (0.8 ton).
Z = (54.5 - 55) / 0.8
Z = -0.625
2. Next, use the Z-score to look up the probability in the standard normal distribution table. The table gives the probability of getting a Z-score less than the given value.
From the table, a Z-score of -0.625 corresponds to a probability of 0.2660.
Therefore, the probability that one car chosen at random will have less than 54.5 tons of coal is approximately 0.2660.
To find the probability that 37 cars chosen at random will have a mean load weight x̄ of less than 54.5 tons of coal, we need to calculate the Z-score for the sample mean using the formula:
Z = (x̄ - μ) / (σ / √n)
where x̄ is the sample mean, μ is still the population mean (55 tons), σ is the population standard deviation (0.8 ton), and n is the sample size (37).
1. Calculate the Z-score:
Z = (54.5 - 55) / (0.8 / √37)
Z = -2.739
2. Using the standard normal distribution table, find the probability associated with a Z-score of -2.739. The table gives the probability of getting a Z-score less than the given value.
From the table, a Z-score of -2.739 corresponds to a probability of 0.0026.
Therefore, the probability that 37 cars chosen at random will have a mean load weight x̄ of less than 54.5 tons of coal is approximately 0.0026.