SOLUTION: I need help with my math today.I need to understand why i got these problems wrong on my quiz...can someone please help me...explain them to me thanks... Problem #1 Life of Lig

Algebra ->  Probability-and-statistics -> SOLUTION: I need help with my math today.I need to understand why i got these problems wrong on my quiz...can someone please help me...explain them to me thanks... Problem #1 Life of Lig      Log On


   



Question 87988: I need help with my math today.I need to understand why i got these problems wrong on my quiz...can someone please help me...explain them to me thanks...
Problem #1
Life of Light Bulbs A certain type of light bulb has an average
life of 500 hours, with a standard deviation of 100 hours. The
length of life of the bulb can be closely approximated by a normal
curve. An amusement park buys and installs 10,000 such
bulbs.

Find the shortest and longest lengths of life for the middle
80% of the bulbs.

Problem #2
Find the percent of the total area under the standard normal curve between each pair of z-scores.
z= 0.64 and z = 2.11


Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Problem #1
Life of Light Bulbs A certain type of light bulb has an average
life of 500 hours, with a standard deviation of 100 hours. The
length of life of the bulb can be closely approximated by a normal
curve. An amusement park buys and installs 10,000 such
bulbs.
Find the shortest and longest lengths of life for the middle
80% of the bulbs.
--------
mu=500; sigma=100
The middle 80% has 40% above the mean and 40% below the mean.
You should draw this picture using a sketch of the normal curve.
and labeling the horizontal axis X.
------
the x-value at the 40% point can be determined by using your
TI calculator or your z-chart. The z-value = InvNorm(0.9)=1.2815
-------
You find the corresponding x-value using the equation z =(x-mu)/sigma
1.2815=(x-500)/100; x-500=128.15; x=628.15 hrs.
That is the upper limit of the middle 80% region.
--------
The lower limit of the 80% region has a z-value of -1.2815
The corresponding x-value is the solution of -1.2815=(x-500)/100
-128.15 = x-500; x=498.72 hrs.
-------------------
Problem #2
Find the percent of the total area under the standard normal curve between each pair of z-scores.
z= 0.64 and z = 2.11
--------
You should use your TI calculator: normalcdf(0.64,2.11)=0.24366
or your z-chart.
=================
Cheers,
Stan H.