Question

Suppose that the amount of time that students spend studying in the library in one sitting is normally distributed with mean 43 minutes and standard deviation 17 minutes. A researcher observed 43 students who entered the library to study. Round all answers to 4 decimal places where possible.

What is the distribution of X?X ~ N(,)

What is the distribution of x¯? x¯ ~ N(,)

What is the distribution of ∑x? ∑x ~ N(,)

If one randomly selected student is timed, find the probability that this student's time will be between 41 and 44 minutes.

For the 43 students, find the probability that their average time studying is between 41 and 44 minutes.

Find the probability that the randomly selected 43 students will have a total study time more than 1978 minutes.

The top 20% of the total study time for groups of 43 students will be given a sticker that says "Great dedication". What is the least total time that a group can study and still receive a sticker? _minutes

Answer 1

here mean time spent studying in libraray in one sitting = = 43 mins

standard deviation = = 17 mins

Here sample size n = 43

standard error of sample mean = 17/sqrt(43) = 2.5925 mins

Herethe distribution of X is N (43, 17)

Here the distribution of sample mean is N (43, 2.5925)

Here the distribution of ∑x is

Mean = 43 * 43 = 1849 mins

Standard deviation = 17 * sqrt(43) = 111.476 mins

so here the distribution of ∑x is 111.476 mins

Here

if x is the student's time in library then

Pr(41 mins < X < 44 mins) = Pr(x < 44 mins ; 43 mins; 17 mins) - Pr(x < 41 mins ; 43 mins ; 17 mins)

Z₂ = (44 - 43)/17 = 0.0588

Z₁ = (41 - 43)/17 = - 0.1176

Pr(41 mins < X < 44 mins) = Pr(Z < 0.0588) - Pr(Z < - 0.1176)

= 0.5234 - 0.4532

= 0.0703

Here for sample mean to lie between 41 and 44 minutes

Pr(41 mins < < 44 mins) = Pr( < 44 mins ; 43 mins ; 2.5925 mins) - Pr( < 41 mins ; 43 mins ; 2.5925 mins)

Z₂ = (44 - 43)/2.5925 = 0.3857

Z₁ = (41 - 43)/17 = - 0.7715

Pr(41 mins < X < 44 mins) = Pr(Z < 0.3857) - Pr(Z < - 0.7715)

= 0.6502 - 0.2202 = 0.43

Here for summation of total study time is

Pr(∑x > 1978 minutes) = 1 - Pr(∑x< = 1978 ) = 1 - Pr(∑x < 1978 ; 1849 mins ; 111.476 mins)

Z = (1978 - 1849)/111.476 = 1.1572

Pr(∑x > 1978 minutes) = 1 - Pr(∑x< = 1978 ) = 1 - Pr(∑x < 1978 ; 1849 mins ; 111.476 mins)  

= 1 - Pr(Z < 1.1572)

= 1 - 0.8764 = 0.1236

(4) Here the least total time a group can study is ∑x₀

Pr(∑x < ∑x₀​) = NORM(∑x < ∑x₀​​ ; 1849 mins ; 111.476 mins) = 0.20

Here the respective Z value is = 0.84162

so,

( ∑x₀​​ - 1849)/111.476 = 0.84162

∑x₀​​ = 1849 + 111.476 * 0.84162 = 1942.82 mins or 1943