<PRE><B>Question 654345</B>
Assume that the weekly demand for Ford car sales follows a normal distribution with mean 50,000 cars and standard deviation 14,000 cars.
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a) There is a 1% chance that Ford will sell more than what number of cars during the next year?
Find the z-value with a left tail of 99%: z = 2.3263
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Find the corresponding x(car sales number) using x = z*s+u
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# of car sales = 2.3263*14000+50000 = 82569
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b) What is the probability that Ford will sell between 2.4 and 2.7 million cars during the next year?
Find the z-values for 2.4 mil and 2.7 mil ; then find the probability
that lies between those z-values.
 
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Cheers,
Stan H.
</PRE>