Question

Assume that the national average score on a standardized test is 1010, and the standard deviation is 200, where scores are normally distributed. What is the probability that a test taker scores at least 1600 on the test? Round your answer to two decimal places.

Answer 1

Answer:

The probability of a test taker scoring at least 1600 is equal to 0.00164.

Step-by-step explanation:

Approach 1 using Normal Distribution Tables:

As we know that for normal distribution z(x) = (x-Mu)/SD - Equation 1

Given Data:

Average scores (Mean) = Mu = 1010

Standard Deviation of scores = SD = 200

What we have to find out:

The probability of having scores at least 1600 will be equal to probability of having scores 1600 and more i.e.

P(x>=1600)

Moreover as we know that P(x<=1599) + P(x>=1600) = 1

Therefore: P(x>=1600) = 1- P(x<=1599)  ; Equation 2

Therefore to calculate P(x<=1599) we have x = 1599

and using equation 1 we have: z(x) = z(1599) = (1599-1010)/200

z(1599) = 0.99836

Then using equation 2 we get:

P(x>=1600) = 1 - 0.99836

P(x>=1600) = 0.00164

Approach 2 using Excel or Google Sheets:

Probability of having scores less than or equal to 1599 will be found out by = 1 - norm.dist(1599,Mu,SD,Commutative)

Probability of having scores less than or equal to 1599 = 1 - norm.dist(135,1010,200,1)

and then again using equation 2 we can find out the probability of scoring at least 1600

PS: Standard normal distribution tables are being attached for reference.

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Answer 2

Answer:

0.00

Step-by-step explanation:

If the national average score on a standardized test is 1010, and the standard deviation is 200, where scores are normally distributed, to calculate the probability that a test taker scores at least 1600 on the test, we should first to calculate the z-score related to 1600. This z-score is $z=\frac{1600-1010}{200}=2.95$, then, we are seeking P(Z > 2.95), where Z is normally distributed with mean 0 and standard deviation 1. Therefore, P(Z > 2.95) = 0.00