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Answer:
464 students will score between 48 and 75. Using the z-distribution, we measure how many standard deviations each score is from the mean, then find the p-value associated with each score to find the proportion, and from the proportion, we find how many out of 1000.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
It assumes the scores are normally distributed with a mean score of 75 and a standard deviation of 15.
This means that
How many students will score between 48 and 75?
First we find the proportion, which is the pvalue of Z when X = 75 subtracted by the pvalue of Z when X = 48. So
X = 75
has a p-value of 0.5
X = 48
has a p-value of 0.0359
1 - 0.0359 = 0.4641
Out of 1000:
0.4641*1000 = 464
464 students will score between 48 and 75. Using the z-distribution, we measure how many standard deviations each score is from the mean, then find the p-value associated with each score to find the proportion, and from the proportion, we find how many out of 1000.
Answer:
Step-by-step explanation:
From Bayes' theorem is stated mathematically as the following equation:[2]
{\displaystyle P(A\mid B)={\frac {P(B\mid A)\,P(A)}{P(B)}},}
where A and B are events and P(B) ≠ 0.
P(A) and P(B) are the probabilities of observing A and B without regard to each other.
P(A | B), a conditional probability, is the probability of observing event A given that B is true.
P(B | A) is the probability of observing event B given that A is true.
At this point, go through the attached file before you continue with part B.
Part B)
P(silver) = P(silver from SS)+P(silver from GS)
note P(SS)=P(GG)=P(GS) = 1/3
P(silver from SS) = 1
P(silver from GS) = 1/2
hence
P(Silver from SS) = 1/3
P(Silver from GS) = 1/3 *1/2
P(Silver) = 1/3*1+1/3*1/2
required probability = P(Silver from SS)/P(Silver) = 2/3
