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 84 tons of coal into each car. The actual weights of coal loaded into each car are normally distributed, with mean = 84 tons and standard deviation = 1.6 ton. What is the probability that 34 cars chosen at random will have a mean load weight x of less than 83.5 tons of coal? (Round your answer to four decimal places.)
To solve this problem, we can use the Central Limit Theorem which states that the distribution of sample means will be approximately normally distributed, regardless of the shape of the population distribution, as long as the sample size is large enough.
We are given that the population mean is 84 tons and the population standard deviation is 1.6 tons. Since we are sampling 34 cars at random, which is considered a large enough sample size, we can use the normal distribution to approximate the distribution of sample means.
The mean of the sampling distribution of sample means, also known as the mean of x-bar, is equal to the population mean, which is 84 tons.
The standard deviation of the sampling distribution of sample means, also known as the standard error of x-bar, is equal to the population standard deviation divided by the square root of the sample size, which is 1.6 tons / sqrt(34).
We want to find the probability that the mean load weight x is less than 83.5 tons. This can be written as P(x < 83.5).
To calculate this probability, we need to standardize the value of 83.5 using the z-score formula:
z = (x - mean) / standard deviation
z = (83.5 - 84) / (1.6 / sqrt(34))
Calculating this value, we find that z ≈ -1.2274.
Using a standard normal distribution table or a calculator, we can find the probability that z < -1.2274 is approximately 0.1101.
Therefore, the probability that 34 cars chosen at random will have a mean load weight x of less than 83.5 tons of coal is approximately 0.1101.
To solve this problem, we can use the Central Limit Theorem, which states that the distribution of sample means from a population with any distribution will approximate a normal distribution as the sample size increases.
First, we need to calculate the mean and standard deviation of the sample mean load weight.
The mean of the sample mean load weight is equal to the mean of the population, which is 84 tons.
The standard deviation of the sample mean load weight, also known as the standard error, can be calculated using the formula:
Standard Error = standard deviation / sqrt(sample size)
In this case, the standard deviation is 1.6 tons, and the sample size is 34 cars.
Standard Error = 1.6 / sqrt(34)
Next, we need to convert the given value of 83.5 tons to a z-score.
Z = (x - mean) / standard error
Z = (83.5 - 84) / (1.6 / sqrt(34))
Using a z-table or calculator, we can find the probability corresponding to this z-score.
The probability that 34 cars chosen at random will have a mean load weight less than 83.5 tons is approximately 0.0085 (rounded to four decimal places).
To solve this problem, we need to use the concept of the sampling distribution of the sample mean.
The given information tells us that the mean weight of coal loaded into each car is 84 tons with a standard deviation of 1.6 tons. We are interested in the mean load weight of a sample of 34 cars chosen at random, denoted as x.
First, we need to find the standard deviation of the sampling distribution of the sample mean, also known as the standard error. The standard error (SE) can be calculated using the formula:
SE = standard deviation / square root of sample size
In this case, the standard deviation is given as 1.6 tons, and the sample size is 34. Therefore, the standard error is:
SE = 1.6 / √34
Next, we need to find the z-score, which represents the number of standard errors the sample mean is away from the population mean. The z-score is calculated using the formula:
z = (x - μ) / SE
Where:
x = desired mean load weight (83.5 tons)
μ = population mean load weight (84 tons)
SE = standard error (calculated above)
Plugging in the values, we have:
z = (83.5 - 84) / (1.6 / √34)
Simplifying this calculation gives us the z-score.
Finally, we can use a standard normal distribution table or a calculator to find the probability associated with the calculated z-score. The probability represents the likelihood that the mean load weight of the sample is less than 83.5 tons.
Round your answer to four decimal places.