For a certain group of individuals, the mean hemoglobin level in the blood is 21.0 grams per milliliter (g/ml). The standard deviation is 2 g/ml. If a sample of 25 individuals is selected, find the probability that the mean will be greater than 21.3 g/ml. Assume the variable is normally distributed. please input as decimal form
Why did the blood cell go to therapy?
Because it was feeling unstable and needed to work on its mean!
Now, let's calculate the probability. To find the probability that the mean will be greater than 21.3 g/ml, we're going to use the Central Limit Theorem.
According to the Central Limit Theorem, for a large enough sample size (which is considered 30 or more in general), the sample mean will follow a normal distribution even if the population distribution is not normal. With a sample size of 25, we can assume a normal distribution for the sample mean.
We need to calculate the standard error of the mean (SEM) first, which is the standard deviation divided by the square root of the sample size:
SEM = standard deviation / √sample size
= 2 g/ml / √25
= 2 g/ml / 5
= 0.4 g/ml
Now, we can calculate the z-score using the formula:
z = (sample mean - population mean) / SEM
= (21.3 g/ml - 21.0 g/ml) / 0.4 g/ml
= 0.3 g/ml / 0.4 g/ml
= 0.75
To find the probability that the mean will be greater than 21.3 g/ml, we need to find the area under the normal distribution curve to the right of the z-score of 0.75. Using a standard normal distribution table or calculator, we find that the probability is approximately 0.2266.
So, the probability that the mean will be greater than 21.3 g/ml is approximately 0.2266.
To find the probability that the mean hemoglobin level will be greater than 21.3 g/ml, we can use the Z-test.
1. First, we need to calculate the standard error of the mean (SEM), which is the standard deviation divided by the square root of the sample size.
SEM = σ / √n
where σ is the standard deviation and n is the sample size.
In this case, σ = 2 g/ml and n = 25. Hence,
SEM = 2 / √25 = 2 / 5 = 0.4 g/ml
2. Next, we calculate the Z-score using the formula:
Z = (X - μ) / SEM
where X is the desired mean value, μ is the population mean, and SEM is the standard error of the mean.
In this case, X = 21.3 g/ml and μ = 21.0 g/ml. Hence,
Z = (21.3 - 21.0) / 0.4 ≈ 0.75
3. Now, we need to find the probability corresponding to the Z-score using a Z-table or calculator.
The Z-table provides the probability to the left of the Z-score. Since we want the probability that the mean is greater than 21.3 g/ml, we need to find the area to the right of the Z-score.
From the Z-table, the probability corresponding to a Z-score of 0.75 is approximately 0.2266.
4. However, we need to find the area to the right of the Z-score. So, the probability that the mean will be greater than 21.3 g/ml is:
1 - 0.2266 = 0.7734 (approximately)
Therefore, the probability that the mean hemoglobin level will be greater than 21.3 g/ml is approximately 0.7734 when selecting a sample of 25 individuals.
To find the probability that the mean hemoglobin level will be greater than 21.3 g/ml for a sample of 25 individuals, we need to use the Central Limit Theorem. According to this theorem, when the sample size is large enough, the distribution of sample means approximates a normal distribution, regardless of the shape of the population distribution.
First, we need to calculate the standard deviation of the sample mean (also known as the standard error). The standard error (SE) is equal to the standard deviation of the population divided by the square root of the sample size.
SE = (Standard Deviation of Population) / sqrt(Sample Size)
= 2 g/ml / sqrt(25)
= 2 g/ml / 5
= 0.4 g/ml
Next, we need to calculate the z-score, which represents how many standard errors the mean is away from the population mean.
z = (Sample Mean - Population Mean) / Standard Error
= (21.3 g/ml - 21.0 g/ml) / 0.4 g/ml
= 0.3 g/ml / 0.4 g/ml
= 0.75
Now, we need to find the probability of having a z-score greater than 0.75. We can use a standard normal distribution table (also known as the z-table) or a calculator to find this probability.
From the z-table, we can look up the corresponding probability for a z-score of 0.75. The probability is approximately 0.7734.
Therefore, the probability that the mean hemoglobin level will be greater than 21.3 g/ml is 0.7734, which should be input as a decimal form.