The mean survivial time after diagnosis for a certain disease is 15 years with a standard deviation of 5 years. Based on a parti
1 answer:
Answer:
The time the patient expected to survive after diagnosis is 29 years.
Step-by-step explanation:
It is provided that the mean survival time after diagnosis for a certain disease is 15 years with a standard deviation of 5 years.
That is,
An individual's predicted survival time is <em>a</em> = 2.8 standard deviations beyond the mean.
Compute the time the patient expected to survive after diagnosis as follows:
Thus, the time the patient expected to survive after diagnosis is 29 years.
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Answer:
89.44% probability that less than 80% of the sample would report eating healthily the previous day
Step-by-step explanation:
We use the binomial approximation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
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.
When we are approximating a binomial distribution to a normal one, we have that ,
.
In this problem, we have that:
So
What is the approximate probability that less than 80% of the sample would report eating healthily the previous day?
This is the pvalue of Z when . So
has a pvalue of 0.8944
89.44% probability that less than 80% of the sample would report eating healthily the previous day
Answer:
Step-by-step explanation: