Answer:
The lowest score a college graduate must be 577.75 or greater to qualify for a responsible position and lie in the upper 6%.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 500
Standard Deviation, σ = 50
We are given that the distribution of test score is a bell shaped distribution that is a normal distribution.
Formula:
We have to find the value of x such that the probability is 0.06.
P(X > x) = 6% = 0.06
Calculation the value from standard normal z table, we have,
Hence, the lowest score a college graduate must be 577.75 or greater to qualify for a responsible position and lie in the upper 6%.
The answer is: "p = 2" .
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Explanation:
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Given:
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" 8p + 8 = 10 + 7p " ; Solve for "p" ;
Subtract "7p" from each side of the equation; & subtract "8" from each side of the equation; as follows:
8p + 8 − 7p − 8 = 10 + 7p − 7p <span>− 8 ;
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to get:
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"p = 2 " ; which is our answer.
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Correct Ans:Option A. 0.0100
Solution:We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.
First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:
So, 90 converted to z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.
Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.
Answer:
10
Step-by-step explanation:
the range is the difference between the biggest and the smallest number
55-45=10
Answer:
I beleive its 6
Step-by-step explanation:
