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The scores for a particular examination are normally distributed with a mean of 68.5% and a standard deviation of 8.2%. What is the probability that a student who wrote the examination had a mark between 80% and 100%? Give your answer to the nearest hundredth.

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Answer:

[tex]P(80/100<x<100/100)=0.08[/tex]

Step-by-step explanation:

We are given that

Mean,[tex]\mu=68.5[/tex]%=68.5/100

Standard deviation, [tex]\sigma=8.2[/tex]%=8.2/100

We have to find the probability that a student who wrote the examination had a mark between 80% and 100%.

[tex]P(80/100<x<100/100)=P(\frac{80/100-68.5/100}{8.2/100}<\frac{x-\mu}{\sigma}<\frac{100/100-68.5/100}{8.2/100})[/tex]

[tex]P(80/100<x<100/100)=P(1.40<Z<3.84)[/tex]

We know that

[tex]P(a<Z<b)=P(Z<b)-P(Z<a)[/tex]

Using the formula

[tex]P(80/100<x<100/100)=P(Z<3.84)-P(Z<1.40)[/tex]

[tex]P(80/100<x<100/100)=0.99994-0.91924[/tex]

[tex]P(80/100<x<100/100)=0.0807\approx 0.08[/tex]

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