From Z-table, Lookup for Z-value corresponding to area 0.50 to the left of the normal curve.
------------------------------------------------------------------------------------------------------
From Z-table, Lookup for Z-value corresponding to area 0.70 to the right of the normal curve.
------------------------------------------------------------------------------------------------------
From Z-table, Lookup for Z-value corresponding to area 0.425 to the left of the normal curve.
From Z-table, Lookup for Z-value corresponding to area 0.425 to the right of the normal curve.
------------------------------------------------------------------------------------------------------
From Z-table, Lookup for Z-value corresponding to area 0.01 to the left of the normal curve.
The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed...
The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean = 41 and standard deviation = 4. Use the TI-84 Plus calculator to answer the following. (a) Find the 21st percentile of the tire lifetimes. (b) Find the 73rd percentile of the tire lifetimes. (c) Find the first quartile of the tire lifetimes. (d) The tire company wants to guarantee that its tires will last at least a certain number of miles....
The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean =μ39 and standard deviation =σ5. (a) What is the probability that a randomly chosen tire has a lifetime greater than 47 thousand miles? (b) What proportion of tires have lifetimes between 38 and 43 thousand miles? (c) What proportion of tires have lifetimes less than 44 thousand miles? Round the answers to at least four decimal places.
QUESTION 11 The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean N = 43 and standard deviation 0 = 5. What proportion of tires have lifetimes between 40 and 50 thousand miles? O 2.0000 0.3550 1.1935 O 0.6450
Normal Distribution. The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mu = 42 and sigma = 5. What is the probability that a randomly chosen tire has a lifetime greater than 45 thousand miles? For this problem we want just the answer. Please give up to 4 significant decimal places, and use the proper rules of rounding.
Mechanic sells a brand of automobile tire that has a life expectancy that is normally distributed, with a mean life of 27,000 miles and a standard deviation of 2200 miles. He wants to get into a free replacement of tires That don’t wear well. How Should he word his guarantee if he is willing to replace approximately 10% of the tires?Tires that wear out by miles will be replaced free of charge. (Round to the nearest mile as needed.)
A normal population has mean = 9 and standard deviation -5. (a) What proportion of the population is less than 19? (b) What is the probability that a randomly chosen value will be greater than 4? Round the answers to four decimal places. Part 1 of 2 The proportion of the population less than 19 is Part 2 of 2 The probability that a randomly chosen value will be greater than 4 is : A normal population has mean =...
A mechanic sells a brand of automobile tire that has a life expectancy that is normally distributed, with a mean life of 30,000 miles and a standard deviation of 2800 miles. He wants to give a guarantee for free replacement of tires that don't wear well. How should he word his guarantee if he is willing to replace approximately 10% of the tires?
Suppose that the life span of a certain automobile tire is normally distributed with mu equals 23,000 miles and sigmaequals2500 miles. (a) Find the probability that a tire will last between 28,000 and 30,500 miles. (b) Find the probability that a tire will last more than 29,000 miles.
If the lifetime X of a certain kind of automobile battery is normally distributed with mean of 3 years and standard deviation of 1 year, and the manufacturer wishes to guarantee the battery for 2 years, what percentage of the batteries will he have to replace under the guarantee? 5.
23. A mechanic sells a brand of automobile tire that has a life expectancy that is normally distributed, with a mean life of 35,000 miles and a standard deviation of 2800 miles. He wants to give a guarantee for free replacement of tires that don't wear well. How should he word his guarantee if he is willing to replace approximately 10% of the tires?