# Math Help - Data help! Basic Probability!

1. ## Data help! Basic Probability!

1) The weekly demand for laser printer cartridges at Office Oasis is normally distributed with a mean of 350 cartridges and a standard deviation of 10 cartridges. There store has a policy of avoiding stockouts (having no product on hand). The manager decides that she wants the chance of a stockout in any given week to be at max 5%. How many cartridges should the store carry each week to meet this policy?

a: 367

2) On a certain section of highway, 90 percent of the motorists drive speeds greater than 100km/h and only 5 percent of the motorists drive at speeds less than 95km/h. if the speeds are assumed normally distributed,

a) determine the motorists mean driving speed and standard deviation.

b) determine the percentage of motorists that drive at speeds greater than 120 or less( last part is cut off!)

a) mean = 121.67; sd = 16.67 b) 53.9%

2. Originally Posted by duffman2003
1) The weekly demand for laser printer cartridges at Office Oasis is normally distributed with a mean of 350 cartridges and a standard deviation of 10 cartridges. There store has a policy of avoiding stockouts (having no product on hand). The manager decides that she wants the chance of a stockout in any given week to be at max 5%. How many cartridges should the store carry each week to meet this policy?

a: 367
You need to find $P\left(Z<\frac{X-350}{10}\right) = 0.05$

3. I am trying to do this in my head.. but I don't get what to do from this step.. Do I change my 0.05 = z-score which is like -1.5 on the table and then solve for x? I get like 333.5...! Can someone please explain this to me!

4. $P(Z<\frac{X-350}{10}) = 0.05
$

[COLOR=Black]Find 0.05 on the z-table, it should correspond to a value of -1.645

Then:
$\frac{X-350}{10}=-1.645$
Solving for x we get 333.55.

So the store should stock about 334 cartridges if they want to have a 5% chance of running out, which makes sense because when the were carrying the the mean of 350 cartridges, they never ran out.