Question

Data collected over a long period of time show that the length of time x to complete a particular college entrance test is normally distributed with an average of 125 minutes and a standard deviation of 18 minutes. What is the probability that a student taking this test will finish in 100 minutes or less? round your answer to 4 decimal places. Remember to round all z values to 2 decimal places.

Answer 1

Using the z-table, the probability that a student taking this test will finish in 100 minutes or less is 0.0824 or 8.24%.

For a normally distributed set of data, given the mean and standard deviation, the probability can be determined by solving the z-score and using the z-table.

First, solve for the z-score using the formula below.

z-score = (x – μ) / σ

where x = individual data value = 100

μ = mean = 125

σ = standard deviation = 18

z-score = (100 – 125) / 18

z-score = (-25) / 18

z-score = -1.39

Find the probability that corresponds to the z-score in the z-table. (see attached images)

-1.39 - (-1.3) : -1.4 - (-1.3) = x - 0.0968 : 0.0808 - 0.0968

-0.09 : -0.1 = x - 0.0968 : -0.016

x - 0.0968 = -0.09(-0.016)/-0.1

x = -0.0144 + 0.0968

x = 0.0824

x = 0.0824

Hence, the probability that a student taking this test will finish in 100 minutes or less is 0.0824 or 8.24%.

Learn more about probability here: brainly.com/question/26822684

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