The Eagles basketball team scored 12 more than 2 times as many points in the last game of the season than in the first game. Whi
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Answer:
39.17% probability that a woman in her 60s who has a positive test actually has breast cancer
Step-by-step explanation:
Bayes Theorem:
Two events, A and B.
In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.
In this question:
Event A: Positive test.
Event B: Having breast cancer.
3.65% of women in their 60s get breast cancer
This means that
A mammogram can typically identify correctly 85% of cancer cases
This means that
Probability of a positive test.
85% of 3.65% and 100-95 = 5% of 100-3.65 = 96.35%. So
What is the probability that a woman in her 60s who has a positive test actually has breast cancer?
39.17% probability that a woman in her 60s who has a positive test actually has breast cancer
<u>Answer:</u>
The probability of a randomly selected student scoring in between 77.6 and 88.4 is 0.8185.
<u>Solution:</u>
Given, Scores on a test are normally distributed with a mean of 81.2
And a standard deviation of 3.6.
We have to find What is the probability of a randomly selected student scoring between 77.6 and 88.4?
For that we are going to subtract probability of getting more than 88.4 from probability of getting more than 77.6
Now probability of getting more than 88.4 = 1 - area of z – score of 88.4
So, probability of getting more than 88.4 = 1 – area of z- score(2)
= 1 – 0.9772 [using z table values]
= 0.0228.
Now probability of getting more than 77.6 = 1 - area of z – score of 77.6
So, probability of getting more than 77.6 = 1 – area of z- score(-1)
= 1 – 0.1587 [Using z table values]
= 0.8413
Now, probability of getting in between 77.6 and 88.4 = 0.8413 – 0.0228 = 0.8185
Hence, the probability of a randomly selected student getting in between 77.6 and 88.4 is 0.8185.