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Answer :
100% surgical procedures would fall within the range of 97.6 and 125.6.
First, let's calculate the z-scores for the lower and upper bounds:
For 97.6 minutes:
[tex]\[ z_1 = \frac{{97.6 - 111.6}}{{2.8}} \][/tex]
[tex]\[ z_1 = \frac{{-14}}{{2.8}} = -5 \][/tex]
For 125.6 minutes:
[tex]\[ z_2 = \frac{{125.6 - 111.6}}{{2.8}} \][/tex]
[tex]\[ z_2 = \frac{{14}}{{2.8}} = 5 \][/tex]
Now, we'll use a standard normal distribution table or a calculator to find the area under the curve between these z-scores.
However, since the z-scores are symmetric around the mean in a standard normal distribution, we only need to find the area between 0 and 5, as the area from -5 to 0 is the same.
From the standard normal distribution table or calculator, the area to the left of z = 5 is almost 1 (0.999999999998), and the area to the left of z = 0 is 0.5.
So, the area between 0 and 5 is approximately 1 - 0.5 = 0.5.
Since this is a symmetric distribution, the area between -5 and 0 is also 0.5.
Therefore, the total area between -5 and 5 is 0.5 + 0.5 = 1, which corresponds to 100%.
Thus, 100% of the surgical procedures fall within the range of 97.6 and 125.6 minutes.
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Final answer:
You would use the formula for calculating z-scores and then use the z-table or a statistical software to find the percentage of surgical procedures that take anywhere between 97.6 and 125.6 minutes, assuming a perfectly normal distribution.
Explanation:
In this problem, we are dealing with a normal distribution, where the mean of the surgical procedure time is 111.6 minutes and the standard deviation is 2.8 minutes. The question asks for the percentage of surgical procedures that fall between 97.6 and 125.6 minutes.
To begin the calculations, you would first need to convert the times into z-scores using the formula: z=(X-μ)/σ where X is the raw score, μ is the mean, and σ is the standard deviation. For 97.6 minutes, the z-score would be (97.6-111.6)/2.8, and for 125.6 minutes, the z-score would be (125.6 - 111.6)/2.8.
Next, you'd use a z-table or a statistical software to find the probability associated with these z-scores (which corresponds to the percentage of surgical procedures falling within these times), subtracting the lower from the higher.
In the context of this problem, it is important to note that the results are based on the assumption of a perfectly normal distribution and actual data may display some amount of variation.
Learn more about normal distribution here:
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