document.write( "Question 853777: 3. The lifetimes of a colour TV picture tube is normally distributed, with a mean of 8 yrs and a standard deviation of 2 years.
\n" ); document.write( "(i) What is the probability that a picture tube will last more than 10 yrs?
\n" ); document.write( "(ii) If the firm guarantees the picture tube for 4 yrs, what percentage of the tubes sold will have to be replaced?
\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #514308 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
\n" );
document.write( "Hi,
\n" );
document.write( "mean of 8 yrs and a standard deviation of 2 years.
\n" );
document.write( "(i)Using TI: The syntax is normalcdf(smaller, larger, µ, σ).
\n" );
document.write( "Note: The -9999 is used as the smaller value to be at least 5 standard deviations from the mean.
\n" );
document.write( "P(x > 10) =  1 - normalcdf(-9999,10, 8,2)
\n" );
document.write( "(ii) 
\n" );
document.write( "P(x < 4) = normalcdf(-9999,4, 8,2)
\n" );
document.write( "0r
\n" );
document.write( "P(P(x > 10) = 1 - NORMSDIST(1) =  1 - .8413 = .1587
\n" );
document.write( "P(x < 4) =  =  NORMSDIST(-2) = .0028     .28% will have to be replaced \n" );
document.write( "
\n" );