The experience of a certain hospital showed the the distribution of length stay of patients is normal with a mean of 11.5 days and a standard deviation of 2 days.

a. What percentage of the patients stayed nine days or less?
b.If a new method of nursing care is to be administered to the middle 96% of the group how long should a patient stay to be included in the study?

To solve these questions, we can use the properties of the normal distribution and the Z-score.

a. To find the percentage of patients who stayed nine days or less, we need to find the Z-score of nine and use the standard normal distribution table.

The formula for Z-score is:
Z = (X - μ) / σ

Where:
Z is the Z-score,
X is the value we want to find the percentage for,
μ is the mean, and
σ is the standard deviation.

Let's calculate the Z-score for nine days:
Z = (9 - 11.5) / 2

Z = -2.5 / 2

Z = -1.25

Now, we can use the standard normal distribution table or a calculator to find the percentage associated with a Z-score of -1.25. Looking it up in the table or using a calculator, we find that the percentage is approximately 0.1093 or 10.93%.

Therefore, approximately 10.93% of the patients stayed nine days or less.

b. To find the cutoff point for the middle 96% of the distribution, we need to find the Z-scores corresponding to the areas in the tails.

Since the middle 96% is split into two equal tails (48% in each tail), we need to find the Z-score that corresponds to the area of 48%. We can look for the Z-score in the standard normal distribution table or use a calculator.

Looking it up, we find that a Z-score of approximately 1.75 corresponds to an area of 0.4801 (48%). However, since we need to include both tails, we multiply this by 2:
2 * 0.4801 = 0.9602

So, the Z-score for the middle 96% is approximately 1.75.

Now, we can use the Z-score formula to find the value (patient stay) for this Z-score:
Z = (X - μ) / σ

Substituting the given values:
1.75 = (X - 11.5) / 2

Multiplying both sides by 2:
3.5 = X - 11.5

Adding 11.5 to both sides:
X = 15

Therefore, a patient should stay approximately 15 days to be included in the study for the new nursing care method.

To answer these questions, we will use the properties of the normal distribution.

a. What percentage of the patients stayed nine days or less?
To find the percentage of patients who stayed nine days or less, we need to calculate the area under the normal distribution curve up to the value of nine days.

To do this, we'll use the Z-score formula: Z = (X - μ) / σ, where X is the given value, μ is the mean, and σ is the standard deviation.

First, we calculate the Z-score for nine days:
Z = (9 - 11.5) / 2 = -1.25

Next, we need to find the area to the left of this Z-score. We can use a standard normal distribution table or a calculator. For this example, let's use an online calculator.

Using the calculator, we find that the area to the left of Z = -1.25 is approximately 0.1056 or 10.56%.

Therefore, approximately 10.56% of the patients stayed nine days or less.

b. If a new method of nursing care is to be administered to the middle 96% of the group, how long should a patient stay to be included in the study?
To find the value of X that corresponds to the middle 96% of the distribution, we need to find the Z-scores that enclose this middle area.

Since the distribution is symmetric, we can find the corresponding Z-scores by finding the area to the left and right of the middle 96%, which is (100% - 96%) / 2 = 2%.

Using an online calculator or a standard normal distribution table, we find the Z-score for the left boundary is approximately -2.05 and the Z-score for the right boundary is approximately 2.05.

Now, we can use the Z-score formula to find the corresponding X-values:
For the left boundary:
Z = (X - μ) / σ
-2.05 = (X - 11.5) / 2

Solving for X, we get:
X = -2.05 * 2 + 11.5 = 7.4

For the right boundary:
Z = (X - μ) / σ
2.05 = (X - 11.5) / 2

Solving for X, we get:
X = 2.05 * 2 + 11.5 = 15.6

Therefore, for a patient to be included in the study with the new nursing care method, they should stay between approximately 7.4 days and 15.6 days.

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