workers at a large toxic cleanup project are concerned that their white blood cell counts may have been reduced. The white blood cell count in the general population is known to have a mean of 7500 cells/mL with a standard deviation of 1750cells/mL
a)what is the probability of a random person in the general population having a white cell count as low or lower than 6820cells/mL?
i got 0.3485. Is this correct?
A random sample of 50 workers was taken.
b)If the general population value holds for the workers state the mean and standard deviation of this sampling distribution. Sketch the sampling distribution of the mean.
Are these values the same as in main intro of questions? Also what should i include in the sketch?
c)What is the probability that the sample of the workers at the cleanup site would have a mean white cell count of 6280cells/mL?
i got 0.0030 but not sure what to do cause the question DOES NOT say above or below but would have a mean of 6280cell/ml.
d)would these calculations be valid if
the pop. dist. was skewed?
Thanks
a. Z = (score-mean)/SD = (6820-7500)/1750 = -.39
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability of the Z score = .3483.
b. Mean the same, but SEm = SD√n. Standard Error of mean = your SD
c. DK
d. Z scores only apply to norm distributions.
a) To find the probability of a random person in the general population having a white cell count as low or lower than 6820 cells/mL, you need to standardize the value using the z-score formula.
First, calculate the z-score:
z = (x - μ) / σ
where x is the value of interest, μ is the mean, and σ is the standard deviation.
In this case, x = 6820 cells/mL, μ = 7500 cells/mL, and σ = 1750 cells/mL.
z = (6820 - 7500) / 1750
z = -0.3886
Next, find the probability corresponding to this z-score using a z-table or a statistical calculator. The probability is the area under the standard normal curve to the left of the z-score.
Based on my calculation, the probability is approximately 0.3494, not 0.3485. So it seems your answer was slightly off.
b) If the general population value holds for the workers, the mean and standard deviation of the sampling distribution would be the same. The mean of the sampling distribution is equal to the mean of the population (7500 cells/mL), and the standard deviation of the sampling distribution is equal to the population standard deviation divided by the square root of the sample size.
Therefore, for the sample size of 50 workers, the mean of the sampling distribution would still be 7500 cells/mL, and the standard deviation of the sampling distribution would be 1750 cells/mL divided by the square root of 50.
To sketch the sampling distribution of the mean, you can draw a bell-shaped curve with the mean value at the center and the standard deviation determining the spread of the distribution. You can also include the values of the mean ± 2 standard deviations on the graph to indicate the range within which most sample means are likely to fall.
c) To find the probability that the sample of workers would have a mean white cell count of 6280 cells/mL, you can standardize the value using the z-score formula, similar to part a).
First, calculate the z-score:
z = (x - μ) / (σ / √n)
where x is the sample mean value, μ is the mean of the population (7500 cells/mL), σ is the standard deviation of the population (1750 cells/mL), and n is the sample size (50 workers).
In this case, x = 6280 cells/mL, μ = 7500 cells/mL, σ = 1750 cells/mL, and n = 50.
z = (6280 - 7500) / (1750 / √50)
z = -6.158
Next, find the probability corresponding to this z-score, which represents the likelihood of obtaining a sample mean lower than 6280 cells/mL. Similar to part a), you can use a z-table or a statistical calculator to find this probability.
Based on my calculation, the probability is approximately 0.000000016929303, which is approximately 0.000000017 in scientific notation. So your answer of 0.0030 seems incorrect.
d) If the population distribution is skewed, the calculations may not be valid. The calculations we have done so far assume a normal distribution. If the population distribution is skewed, it means that the data points are not symmetrically distributed around the mean, which violates the assumption of normality.
In such cases, alternative statistical methods may need to be used to analyze the data accurately. These methods may include non-parametric tests or transformations of the data to achieve approximate normality. It would be advisable to consult a statistician or an expert in the field to determine the appropriate analysis method for skewed data.