Answer to Question #327582 in Statistics and Probability for Khaita

Question

Question 4 [25]

The amount of time devoted to preparing for a statistics examination by students is a normally

distributed random variable with a mean of 17 hours and a standard deviation of 5 hours.

Required:

a) What is the amount of time below which only 15% of all students spend studying?

b) What is the amount of time above which only one third of all students spend studying?

c) What is the probability that a student spends between 16 and 20 hours studying?

d) What is the probability that a student spends at least 15 hours studying?

e) What is the probability that a student spends at most 18 hours studying?

Answer 1

$a:\\P\left( X\leqslant C \right) =0.15\Rightarrow P\left( \frac{X-17}{5}\leqslant \frac{C-17}{5} \right) =0.15\Rightarrow \\\Rightarrow \frac{C-17}{5}=z_{0.15}\Rightarrow C=5z_{0.15}+17=5\cdot \left( -1.0364 \right) +17=11.818\\b:\\P\left( X>C \right) =0.3333\Rightarrow P\left( \frac{X-17}{5}\leqslant \frac{C-17}{5} \right) =0.6667\Rightarrow \\\Rightarrow \frac{C-17}{5}=z_{0.6667}\Rightarrow C=5z_{0.6667}+17=5\cdot 0.4308+17=19.154\\c:\\P\left( 16\leqslant X\leqslant 20 \right) =P\left( \frac{16-17}{5}\leqslant \frac{X-17}{5}\leqslant \frac{20-17}{5} \right) =\\=\varPhi \left( 0.6 \right) -\varPhi \left( -0.2 \right) =0.305\\d:\\P\left( X\geqslant 15 \right) =1-P\left( X<15 \right) =1-P\left( \frac{X-17}{5}<\frac{15-17}{5} \right) =\\=1-\varPhi \left( -0.4 \right) =1-0.3446=0.6554\\e:\\P\left( X\leqslant 18 \right) =P\left( \frac{X-17}{5}\leqslant \frac{18-17}{5} \right) =\varPhi \left( 0.2 \right) =0.5793$