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Students who take statistics courses have a normal distribution period of 55 minutes on average and...

Students who take statistics courses have a normal distribution period of 55 minutes on average and 15 minutes of standard deviation. Find the probability that any student spends 65-85 minutes on the exam.

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Solution-

According to question, students who take statistics courses have a normal distribution period of 55 minutes on average and 15 minutes of standard deviation.

So, mean = \mu = 55 minutes

And Standard deviation = \sigma = 15 minutes.

If the random normal variable is X.Then

The probability that any student spends 65-85 minutes

= P( 65 < X < 85)

=P\left ( \frac{65-\mu}{\sigma}<z<\frac{85-\mu}{\sigma} \right )

=P\left ( \frac{65-55}{15}<z<\frac{85-55}{15} \right )

=P\left ( \frac{10}{15}<z<\frac{30}{15} \right )

=P\left ( 0.667<z<2 \right )

=P\left ( 0<z<2 \right )-P(0<z<0.667)

= 0.4772 - 0.2486

= 0.2286

(Values are taken from standard normal distribution table).

Hence, Probability that any student spends 65 minutes to 85 minutes is 0.2286 .

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