Question 273759: I was wondering if this is right :)
4. When a pizza restaurant’s delivery process is operating effectively, pizzas are delivered in an average of
45 minutes with a standard deviation of 6 minutes. To monitor its delivery process, the restaurant randomly
selects five pizzas each night and records their delivery times. For the sake of the argument, assume that the
population of all delivery times on a given evening is normally distributed with a mean of µ = 45 minutes and a standard deviation of δ = 6 minutes.
z= X - mean/standard deviation = 5-45/6=-6.66
The area between z=0 and z=-6.66-= how do i find the area?? do I double 3?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 4. When a pizza restaurant’s delivery process is operating effectively, pizzas are delivered in an average of 45 minutes with a standard deviation of 6 minutes.
To monitor its delivery process, the restaurant randomly selects five pizzas each night and records their delivery times.
For the sake of the argument, assume that the population of all delivery times on a given evening is normally distributed with a mean of µ = 45 minutes and a standard deviation of δ = 6 minutes.
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Comment: You need to be given the mean delivery time for the t pizzas.
Then you need to find the t-value for that sample mean time using
t(x-bar) = (xbar-45)/[6/sqrt(5)]
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Comment: IF the alternate hypothis is "u is not equal to 45", the p-value
for your problem is 2*P(t > the t-value you obtained for x-bar)
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z= (X - mean)/standard deviation = 5-45/6=-6.66
Comment (5-45) is incorrect. 5 is the sample size; 45 is the expected
dilivery time. You need t(x-bar-45) where x-bar is the sample's mean
time.
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The area between z=0 and z=-6.66-= how do i find the area?? do I double
Comment: I'm not sure where you are going with with question. You don't
need the area (which is 0.5000). The doubling is associated with a p-score
for a two-tail test.
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Let me know if you need more help on this problem.
Cheers,
Stan H.
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