Question

In a university class that includes 120 students, mostly non-traditional, the Mean age is 32, with a standard deviation 12 students, and with a normal distribution. Six of these students are international students What is the probability that, if one student is picked at a random, he/she be a teenager (12 to 19 years old)? You need to (somewhat) critically think about this before tackling it!! 1. What is the probability that if ONE student is selected at random, he/she will be an international student? 2. 3. What is the probability that if ONE student is selected at random, his/her age be between 28 to 30 years old? What is the probability that if on student is selected at random, his/her age be between 28 to 36 years old? 4. 5. Suppose we select, at random, a sample of 9 students from this class. What is the probability that the MEAN age of that sample will be more than 38 years old?

Answer 1

Solution :-

Given data :-

Number of students = 120

Mean age, $\mu$ = 32

standard deviation, s = 12

( 1 ) :-

Here we need to find out the probability that, if one student is picked at a random, he/she be a teenagers ( 12 to 19 years old).

i.e, $P(12\leq X\leq 19)$

$P(12\leq X\leq 19)=P(X\leq 19)-P(X\leq 12)$

$P(12\leq X\leq 19)=P\left (\frac{19-\mu }{s} \right )-P\left (\frac{12-\mu }{s} \right )$

$P(12\leq X\leq 19)=P\left (\frac{19-32 }{12} \right )-P\left (\frac{12-32 }{12} \right )$

$P(12\leq X\leq 19)=P\left (\frac{13}{12} \right )-P\left (\frac{-20 }{12} \right )$

$P(12\leq X\leq 19)=P\left (-1.083 \right )-P\left (1.667 \right )$

$P(12\leq X\leq 19)=0.14007-0.04846$ ( $\because$ From standard normal table)

$P(12\leq X\leq 19)=0.09161$      

( 2 ) :-

Here we need to find out the probability that if ONE student is selected at random, he/she will be an international student.

i.e, P(International student )

P(International student ) = 6/120

P(International student ) = 0.05

( 3 ) :-

Here we need to find out the probability that if on student is selected at random, his/her age be between 28 to 30 years old.

i.e, $P(28\leq X\leq 30)$

$P(28\leq X\leq 30)=P\left (z<\frac{30-\mu}{s} \right ) -P\left ( z<\frac{28-\mu}{s} \right )$

$P(28\leq X\leq 30)=P\left (z<\frac{30-32}{12} \right ) -P\left ( z<\frac{28-32}{12} \right )$

$P(28\leq X\leq 30)=P\left (z<\frac{-2}{12} \right ) -P\left (z< \frac{-4}{12} \right )$

$P(28\leq X\leq 30)=P\left (z<-0.1667 \right ) -P\left ( z<-0.333 \right )$

$P(28\leq X\leq 30)=0.43644 -0.37070$    ( $\because$ From standard normal table)

$P(28\leq X\leq 30)=0.06574$

( 4 ) :-

Here we need to find out the probability that if on student is selected at random, his/her age be between 28 to 36 years old.

i.e, $P(28\leq X\leq 36)$

$P(28\leq X\leq 36)=P(X\leq 36)-P(X\leq 28)$

$P(28\leq X\leq 36)=P\left (z<\frac{36-\mu }{s} \right )-P\left (z<\frac{28-\mu }{s} \right )$

$P(28\leq X\leq 36)=P\left (z<\frac{36-32 }{12} \right )-P\left (z<\frac{28-32 }{12} \right )$

$P(12\leq X\leq 19)=P\left (z<\frac{4}{12} \right )-P\left (z<\frac{-4 }{12} \right )$

$P(12\leq X\leq 19)=P\left (z<0.334 \right )-P\left (z<-0.334 \right )$

$P(12\leq X\leq 19)=0.62930 -0.37070$    ( $\because$ From standard normal table)

$P(12\leq X\leq 19)=0.2686$

( 5 ) :-

Here we need to find out the probability that the mean age of that sample will be more than 38 years old.

i.e, $P(x\geq 38)$

$P(x\geq 38)=P\left (z>\frac{38-\mu }{s/\sqrt{n}} \right )$

$P(x\geq 38)=P\left (z>\frac{38-32 }{12/\sqrt{9}} \right )$

$P(x\geq 38)=P\left (z>\frac{6 }{4} \right )$

$P(x\geq 38)=P\left (z>1.5 \right )$

$P(x\geq 38)=1-P\left (z<1.5 \right )$

$P(x\geq 38)=1-0.9332$    ( $\because$ From standard normal table)

$P(x\geq 38)=0.0668$​​​​​​​