The scores on an entrance exam to a university are known to have an approximately normal distribution with mean 65% and standard deviation 7.1%. Using the normalcdf function on your graphing calculator, what percentage of students would score 70 or better on this entrance exam? A. 28.4%
B. 18.9%
C. 24.1%
D. 22.3%

The correct answer is - C. 24.1%

Step-by-step explanation:

Given:

mean μ = 65%

standard deviation δ = 7.1 %

solution:

Prob( X>70) = 1 - Prob(x<70)  

= P (x-μ/δ ≥ 70 -65/7.1)

= 1 - Prob( (70-65)/7.1)

= 1 - Prob ( z < 0.7042553)

= 0.24065

the percentage of students scoring 70 or more in the exam

= 24.065*100

= 24.1%

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i believe it is b and c. (side note, i got the same computer from k-12 lol i’m in 8th grade and this is my first year, will it get

step-by-step explanation: