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skye2598
25.04.2020 •
Mathematics
5. Which table of values represents y = 1.5x + 3
20
LY
I
33
+126
21
11
3
7.5
24
39
12
x
Y
I
I
4
16
11
22
36
9
I
|
8
25.5
|
6
19.5
2
6
T
Solved
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Ответ:
0.2159 = 21.59% probability this student’s score will be at least 1500.
Step-by-step explanation:
To solve this question, we need to understand the normal distribution and conditional probability.
Normal Probability Distribution:
Problems of normal distributions 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.
Conditional Probability
We use the conditional probability formula to solve this question. It is
In which
P(B|A) is the probability of event B happening, given that A happened.
P(A) is the probability of A happening.
In this question:
Event A: Recognized student(scored more than 1350)
Event B: Score of at least 1500.
SAT scores (out of 1600) are distributed normally with a mean of 1100 and a standard deviation of 200
This means that![\mu = 1100, \sigma = 200](/tpl/images/1179/4724/6df5c.png)
Probability of being recognized.
1 subtracted by the pvalue of Z when X = 1350. So
1 - 0.8944 = 0.1056
So![P(A) = 0.1056](/tpl/images/1179/4724/9e5ec.png)
Probabibility of being recognized and scoring at least 1500.
Intersection between more than 1350 and more than 1500 is more than 1500. So this probability is 1 subtracted by the pvalue of Z when X = 1500.
1 - 0.9772 = 0.0228
So,![P(A \cap B) = 0.0228](/tpl/images/1179/4724/81228.png)
What is the probability this student’s score will be at least 1500?
0.2159 = 21.59% probability this student’s score will be at least 1500.