A sample of 100 dry battery cells tested
to find the length of life produced the
following results. Assuming that the
data are normally distributed, find
percentage of battery cells
expressed to have a life of less than 6
hours.
are
(Given that A(2) = 0.4772, Mean = 12
hours and Standard Deviation = 3 hours)
Answers
Answer:
Solution:
We have Meanu - 12 hours, standard deviation=o=3 hours, Total number of objects = N = 100
P(life of battery more than 15 hours)
The standardized value of x 15 is z=(x-μ)/o
(15-12)/3 = 3/3 = 1
P(Z >1)= area under the standard normal curve to the right of z = 1
P(Z >1)= (Area to the right of z=0) - (Area between z=0 and z = 1)
P(Z >1) 0.5-0.3413= 0.1587
P(life of battery less than 6 hours)
The standardized value of x = 6 is z = (x-μ)/o
(6-12)/3 =-6/3=-2
P(Z <-2)= area under the standard normal curve to the
left of 2 = -2
> 1) = (Area to the right of z=0) - (Area between z=0 and z = 1)
P(Z >1)=0.5-0.3413= 0.1587
P(life of battery less than 6 hours)
The standardized value of x = 6 is z = (x-μ)/o
= (6-12)/3 =-6/3= -2
P(Z <-2) = area under the standard normal curve to the
left of z = -2
P(Z <-2)=(Area to the left of z=0)-(Area between z=0
and z= -2)
P(Z >1)= 0.5-0.4772 = 0.0228
P(life of battery between 10 hours and 14 hours)
The standardized value of x = 10 is z=(x-μ)/o
(10-12)/3-2/3=-0.67
The standardized value of x = 14is z=(x-μ)/a
= (14-12)/3 = 2/3 = 0.67
P(-0.67 z<0.67)= area under the standard normal
curve between
z=0.67 and z=-0.67
P(-0.67<z<0.67) = (Area between z = 0 and z=-0.67) +(Area between z = 0 and z = 0.67)
P(-0.67 <z<0.67)=2 x (Area between z = 0 and z = -0.67)
= 2 x 0.2486= 0.4972
Ans: 15.87% batteries have life more than 15 hours.
2.28% batteries have life less than 6 hours.
49.72% batteries have life between 10 hours and 14 hours.
Given : A sample of 100 dry battery cells tested to find the length of life produced the following results.
data are normally distributed,
Mean = 12 Standard Deviation = 3 hours
To find : percentage of battery cells expected to have a life of less than 6
hours.
Solution :
z score = ( Value - Mean )/ SD
Value = 6 hrs
Mean = 12 hrs
SD = 3 hrs
=> z score = ( 6 - 12 )/ 3
=> z score = - 2
Proportion below is 0.0228
Hence 2.28 % of battery cells expected to have a life of less than 6
hours.
or (0.5 - 0.4772 = 0.0228 , 2.28 %)
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