Course MTH 1210
Homework #5
(
answers
)
(
The Normal Distribution
)
Mar 24’09
Use the MS Excel program
Three_distributions.xls,
sheet 3
1.
A certain type of light bulb has an
average life
of
500 hr
, with a
standard deviation
of
100 hr
.
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 total number that can be expected to last for each of
the following periods of time:
(a)
less than 500 hr,
(b)
between 290 and 540 hr,
(c)
more than 300
hr.
Solution.
In this problem we are given that the mean (average) µ = 500 hr and the standard
deviation σ = 100 hr. Run the program
Three_distributions.xls,
sheet
3
and enter the values of
500
and
100
in the cells
G58
and
G59
respectively (type in each value and press [ENTER] key).
We’ll work with the rectangular block of cells
E53 – K62
.
(a)
To answer this question, we should evaluate the product
10,000*f(X ≤ 500)
where
f(X ≤ 500)
is the cumulative probability that the length of life of a bulb is less than
or equal to 500 hr. Enter the value of 500 in the cell
G58
for
x
. We find that
f(X ≤ x) = 0.5
(see the cell
I59
). Thus,
10,000*f(X ≤ 500) =
10,000*0.5
= 5000
.
x =
500
f(x) =
0.003989
µ =
500
σ =
100
f(X ≤ x) =
0.5
(b)
The number in question is
10,000*f(290 ≤ X ≤ 540) = 10,000*[ f(X ≤ 540) – f(X ≤ 290) ].
To evaluate the cumulative probability
f(X ≤ 540)
that the length of life of a bulb is less
than or equal to 540 hr, we enter the value of
540
in the cell
G58
for
x
and find in the cell
I59
that
f(X ≤ x) =
0.655422
.
x =
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 Spring '05
 Kzaer
 Normal Distribution, Probability, Standard Deviation, blood clotting

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