Find the cube root of 19683 by prime factorization method. Show steps.
2 answers:
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Answer:
1) 22.66%
2) 20
Step-by-step explanation:
The scores of a test are normally distributed.
Mean of the test scores = u = 22
Standard Deviation = = 4
Part 1) Proportion of students who scored atleast 25 points
Since, the test scores are normally distributed we can use z scores to find this proportion.
We need to find proportion of students with atleast 25 scores. In other words we can write, we have to find:
P(X ≥ 25)
We can convert this value to z score and use z table to find the required proportion.
The formula to calculate the z score is:
Using the values, we get:
So,
P(X ≥ 25) is equivalent to P(z ≥ 0.75)
Using the z table we can find the probability of z score being greater than or equal to 0.75, which comes out to be 0.2266
Since,
P(X ≥ 25) = P(z ≥ 0.75), we can conclude:
The proportion of students with atleast 25 points on the test is 0.2266 or 22.66%
Part 2) 31st percentile of the test scores
31st percentile means 31%(0.31) of the students have scores less than this value.
This question can also be done using z score. We can find the z score representing the 31st percentile for a normal distribution and then convert that z score to equivalent test score.
Using the z table, the z score for 31st percentile comes out to be:
z = -0.496
Now, we have the z scores, we can use this in the formula to calculate the value of x, the equivalent points on the test scores.
Using the values, we get:
Thus, a test score of 20 represent the 31st percentile of the distribution.