document.write( "Question 853774: 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?\r
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Algebra.Com's Answer #514309 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
\n" ); document.write( "Hi, Previously Posted
\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)
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\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( "
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