Question

A nationwide study on reaction time is conducted on participants in two age groups. The participants in Group X are less than 40 years old. Their reaction times are normally distributed with mean 0.489 seconds and standard deviation of 0.07 seconds.

a. A person is selected at random from Group X. Find the probability that their reaction time is greater than 0.65 seconds. The participants in Group Y are 40 years or older. Their reaction times are normally distributed

b. The participants in Group Y are 40 years or older. Their reaction times are normally distributed with mean 0.592 seconds and standard deviation σ seconds. The probability that the reaction time of a person in Group Y is greater than 0.65 seconds is 0.396. Find the value of σ

c. In the study, 38 % of the participants are in Group X. A randomly selected participant has a reaction time greater than 0.65 seconds. Find the probability that the participant is in Group X.

d. Ten of the participants with reaction times greater than 0.65 are selected at random. Find the probability that at least two of them are in Group X.

Answer 1

For group X: the reaction time is normally distributed with $\mu=0.489,\sigma= 0.07$

a) the probability that their reaction time is greater than 0.65 seconds =$P(X>0.65)=P(Z>\frac{0.65-0.489}{0.07})=P(Z>2.3)=1-P(Z<2.3)=1-0.9893=0.0107$

b) The probability that the reaction time of a person in Group Y is greater than 0.65 seconds is 0.396.

$P(Y>0.65)=P(Z>\frac{0.65-0.592}{\sigma })=0.396$

$P(Y>0.65)=P(Z>\frac{0.65-0.592}{\sigma })=P(Z>0.2637)=0.396$

$\frac{0.65-0.592}{\sigma }=0.2637$

$\frac{0.65-0.592}{0.2637 }=\sigma$

C) This is a conditional probability

Let B be the time that the reaction time is greater than 0.65

$P(X)=0.38,P(Y)=0.62$

$P(B\mid X)=0.0107$

$P(B\mid Y)=0.3960$

the probability that the participant is in Group X : $P(X\mid B)=\frac{0.38*0.0107}{0.38*0.0107+0.62*0.3960}=0.0163$

D) This can be solved using the binomial distribution:

Probability of success: p = 0.0163

Total number of trials: n= 10

  the probability that at least two of them are in Group X.:

P(X > 2) 1-P(X < 1)

1) = Σ¡oc(0.0163)"(1-0.0163)10-,-0.9890 ーエ P(X-

$P(X\geq2 )=1-0.9890=0.0110$

$\sigma=0.22$

C)