Question

1.)

The time college students spend on the internet follows a Normal distribution. At Johnson University, the mean time is 5.5 hours per day with a standard deviation of 1.1 hours per day.

1. If 100 Johnson University students are randomly selected, what is the probability that the average time spent on the internet will be more than 5.8 hours per day?
   Round to 4 places.  
   
2. If 100 Johnson University students are randomly selected, what is the probability that the average time spent on the internet is between 5.3 hour and 5.8 hours?
   Round to 4 places  
   
3. If 100 Johnson University students are randomly selected, what mean number of hours on the internet per day separated the bottom 33% and top 67% of internet usage for college students?  
   hours per day. Round to 2 places.

2.)

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 10.1 years, and standard deviation of 0.5 years.

If 19 items are picked at random, 3% of the time their mean life will be less than how many years?

Give your answer to one decimal place.

Answer 1

Solution

1)

Given :
n=100
mean=mu=5.5 hour per day
Population standard deviation=sigma=1.1 hour per day
For tha sample of size 100,tha sampling mean is same as the population mean
muX_bar = mu=5.5 and
Sampling standard deviation will be ,
SigmaX_ bar=sigma/✓n
=1.1/✓100
=0.11
P(Xbar>5.8)= 1- P(Xbar<=5.8)
=1- p{[(Xbar - mu_Xbar)/sigma_Xbar] -[(5.8-5.5)/0.11]}
=1-p( z<=2.7273)
=1-p( z<=2.73) .........upto 2 decimal place
=1-0.9968 .........from standard normal table
=0.0032
The probability that average time spent on the internet will be more than 5.8 hours per day is 0.0032
B) P(5.3<Xbar<5.8)=P{[(5.3- 5.5)/0.11]<[(Xbar-muX_bar )/sigma_Xbar]<[(5.8- 5.5)/0.11]}
=P(-1.82<Z< 2.73)
=P(z< 2.73) - P(z< 1.82)
=0.9968-0.0344
=0.9624
The probability that average time spent on the internet between 5.3 to 5.8 hours per day is
0.9624
C) By using the standard normal distribution we can find the mean number of hours on the internet per day will be less than 33% is given as
P (z<z)= 0.33
P(z< -0.4399)=0.33
By using the standard normal table.
z= -0.4399
Now using the z score formula ,we get,
Z= ( Xbar-mu_Xbar )/sigma_Xbar
- 0.4399= (Xbar-5.5)/0.11
xbar= (- 0.4399×0.11)+5.5
=-0.0484+5.5
= 5.4516
The mean number of hours on the internet per day separate the bottom 33% and top 67% of internet usage for college students is 5.4516
2)Given :
n=19
mean=mu=10.1 year
Population standard deviation=sigma=0.5 year
For tha sample of size 19,tha sampling mean is same as the population mean
muX_bar = mu=10.1 year and
Sampling standard deviation will be ,
SigmaX_ bar=sigma/✓n
=0.5/✓19
=0.1147
using the standard normal distribution,
P (z<z)= 0.03
P(z< -1.881)=0.03
By using the standard normal table.
z= -1.881
Now using the z score formula ,we get,
Z= ( Xbar-mu_Xbar )/sigma_Xbar
- 1.881= (Xbar-10.1)/0.1147
xbar= (- 1.881×0.1147)+10.1
= - 0.2157+10.1
= 9.8843
3% of the time there mean life will be less than 9.8843 year