The Time Spent In Days Waiting For A Heart Transplant In Two States
The time spent (in days) waiting for a heart transplant in two states for patients with type A+ blood can be approximated by a normal distribution, as shown in the graph to the right. Complete a & B. The graph shows between 45 and 210 days, with mean (µ) = 126 and standard deviation (σ) = 23.6. What is the shortest time spent waiting for a heart that would still place a patient in the top 10% of waiting times? ____ days (round to two decimals). I will send B’s percentage after A is answered through Chat.
Paper For Above instruction
Determining the shortest waiting time to be in the top 10% of transplant wait times involves understanding the properties of the normal distribution. Given the mean (µ) of 126 days and a standard deviation (σ) of 23.6 days, the task is to find the cutoff point at which 90% of patients have shorter wait times, leaving the top 10% with longer waits. This is essentially finding the 90th percentile of the distribution.
In a normal distribution, percentile points are associated with z-scores, which indicate how many standard deviations an element is from the mean. To find the z-score corresponding to the 90th percentile, statistical tables or software such as calculator functions or statistical software (e.g., SPSS, R, or Excel) can be used. The z-score for the 90th percentile is approximately 1.28.
The formula to convert a z-score into a raw score (X) in the context of a normal distribution is:
X = µ + z * σ
Substituting the known values:
X = 126 + 1.28 * 23.6
Calculating the product:
1.28 * 23.6 ≈ 30.21
Adding this to the mean:
X ≈ 126 + 30.21 ≈ 156.21
Thus, the shortest waiting time for a patient to be in the top 10% is approximately 156.21 days. Since the problem requests rounding to two decimals, the answer is 156.21 days.
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