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Answer :
To solve the problems, we will use the properties of the normal distribution and the concept of the sampling distribution of the sample mean.
- For the ICU patients' heart rates:
Since the sample size is 49, which is sufficiently large, the Central Limit Theorem tells us that the sampling distribution of the sample mean will be approximately normally distributed. The mean of the sampling distribution is the same as the population mean, and its standard deviation, often called the standard error (SE), is given by:
[tex]SE = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{49}} = \frac{12}{7} \approx 1.71[/tex]
Let's calculate the probabilities:
a) Probability that the sample mean heart rate is between 83 and 87 bpm
We want to find [tex]P(83 < \bar{x} < 87)[/tex]. To do this, we convert the raw scores into z-scores using [tex]z = \frac{\bar{x} - \mu}{SE}[/tex].
[tex]z_{83} = \frac{83 - 85}{1.71} \approx -1.17[/tex]
[tex]z_{87} = \frac{87 - 85}{1.71} \approx 1.17[/tex]
Now, we find the probability from standard normal distribution tables or by using a calculator:
[tex]P(-1.17 < z < 1.17) \approx P(z < 1.17) - P(z < -1.17) \approx 0.8790 - 0.1210 = 0.7580[/tex]
b) Probability that the sample mean heart rate is less than 85 bpm
Since 85 bpm is the mean, the probability that [tex]\bar{x} < 85[/tex] is essentially the probability that [tex]z < 0[/tex], which is 0.5, because the standard normal distribution is symmetric about zero.
c) Probability that the sample mean heart rate is at least 86 bpm
We want [tex]P(\bar{x} \geq 86)[/tex]. First, find the z-score for 86:
[tex]z_{86} = \frac{86 - 85}{1.71} \approx 0.58[/tex]
Then calculate the probability:
[tex]P(z > 0.58) = 1 - P(z < 0.58) \approx 1 - 0.7190 = 0.2810[/tex]
- For the general population mean:
We have a population with a mean [tex]\mu = 200[/tex] and [tex]\sigma = 50[/tex], and a sample size of 100.
(a) The expected value of [tex]\bar{x}[/tex]
The expected value of the sample mean is the same as the population mean: [tex]E(\bar{x}) = 200[/tex].
(b) The standard deviation of [tex]\bar{x}[/tex], also known as the standard error, is:
[tex]SE = \frac{\sigma}{\sqrt{n}} = \frac{50}{\sqrt{100}} = \frac{50}{10} = 5[/tex]
(c) Show the sampling distribution of [tex]\bar{x}[/tex]
The sampling distribution of [tex]\bar{x}[/tex] is approximately normal, with mean [tex]\mu = 200[/tex] and standard error [tex]SE = 5[/tex].
(d) What does the sampling distribution of [tex]\bar{x}[/tex] show?
The sampling distribution of [tex]\bar{x}[/tex] shows the distribution of sample means that would be obtained if an infinite number of samples of size 100 were taken from the population. It allows us to make inferences about the population mean and understand the variability expected in the sample means.
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