The expected (mean) life of a particular type of light bulb is 1,000 hours with a standard deviation of 50 hours. the life of this bulb is normally distributed. what is the probability that a randomly selected bulb would last longer than 1150 hours?

For any normal random variable X with mean ÎĽ and standard deviation Ď , X ~ Normal( ÎĽ , Ď ), (note that in most textbooks and literature the notation is with the variance, i.e., X ~ Normal( ÎĽ , Ď² ). Most software denotes the normal with just the standard deviation.) You can translate into standard normal units by: Z = ( X - ÎĽ ) / Ď Moving from the standard normal back to the original distribuiton using: X = ÎĽ + Z * Ď Where Z ~ Normal( ÎĽ = 0, Ď = 1). You can then use the standard normal cdf tables to get probabilities. If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central Limit Theorem. If a sample of size is is drawn from a population with mean ÎĽ and standard deviation Ď then the sample average xBar is normally distributed with mean ÎĽ and standard deviation Ď /âš(n) IYou can translate into standard normal units by: Z = ( X - ÎĽ ) / Ď Moving from the standard normal back to the original distribuiton using: X = ÎĽ + Z * Ď Where Z ~ Normal( ÎĽ = 0, Ď = 1). You can then use the standard normal cdf tables to get probabilities. If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central Limit Theorem. If a sample of size is is drawn from a population with mean ÎĽ and standard deviation Ď then the sample average xBar is normally distributed with mean ÎĽ and standard deviation Ď /âš(n) IYou can translate into standard normal units by: Z = ( X - ÎĽ ) / Ď Moving from the standard normal back to the original distribuiton using: X = ÎĽ + Z * Ď Where Z ~ Normal( ÎĽ = 0, Ď = 1). You can then use the standard normal cdf tables to get probabilities. If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central Limit Theorem. If a sample of size is is drawn from a population with mean ÎĽ and standard deviation Ď then the sample average xBar is normally distributed with mean ÎĽ and standard deviation Ď /âš(n) in You can translate into standard normal units by: Z = ( X - ÎĽ ) / Ď Moving from the standard normal back to the original distribuiton using: X = ÎĽ + Z * Ď Where Z ~ Normal( ÎĽ = 0, Ď = 1). You can then use the standard normal cdf tables to get probabilities. If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central Limit Theorem. If a sample of size is is drawn from a population with mean ÎĽ and standard deviation Ď then the sample average xBar is normally distributed with mean ÎĽ and standard deviation Ď /âš(n) in this question we have X ~ Normal( ÎĽx = 1000 , Ďx² = 2500 ) X ~ Normal( ÎĽx = 1000 , Ďx = 50 ) Find P( X > 1150 ) P( ( X - ÎĽ ) / Ď > ( 1150 - 1000 ) / 50 ) = P( Z > 3 ) = P( Z < -3 ) = 0.001349898 Find P( X < 1100 ) P( ( X - ÎĽ ) / Ď < ( 1100 - 1000 ) / 50 ) = P( Z < 2 ) = 0.9772499

The answer would be: b