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Answer:
a) N(4, 1.6)
b) 4 days
c)
d) There is a 30.85% probability of spending more than 4.8 days in recovery.
e) There is a 11.62% probability of spending between 3.2 and 3.7 days in recovery.
f) The 90th percentile for recovery times is 6.05 days.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
Mean of 4 days and a standard deviation of 1.6 days. So .
a. What is the distribution of X?
This is normal with mean 4 and standard deviation 1.6. So N(4, 1.6).
b. What is the median recovery time?
For the normal distribution, the median is the same as the mean. So the median recovery time is 4 days.
c. What is the Z-score for a patient that took 5.1 days to recover?
d. What is the probability of spending more than 4.8 days in recovery?
This probability is 1 subtracted by the pvalue of Z when . So
has a pvalue of 0.6915.
This means that there is a 1-0.6915 = 0.3085 = 30.85% probability of spending more than 4.8 days in recovery.
e. What is the probability of spending between 3.2 and 3.7 days in recovery?
This probability is the pvalue of Z when X = 3.7 subtracted by the pvalue of Z when X = 3.2. So:
X = 3.7
has a pvalue of 0.4247
X = 3.2
has a pvalue of 0.3085
This means that there is a 0.4247 - 0.3085 = 0.1162 = 11.62% probability of spending between 3.2 and 3.7 days in recovery.
f. The 90th percentile for recovery times is
This probability is the value of X when Z has a pvalue of 0.90. So it is X when Z = 1.28.
The 90th percentile for recovery times is 6.05 days.