Have the mean, standard deviation and probability...but something seems off!

hi all!

So I am taking a Quantitative methods class for my MBA and we are just getting started, so I am somewhat overwhelmed. I am taking the course online so most of it is me interpreting the text. I have a homework question I solved but it's just not sitting right with me. If you wouldn't mind me bouncing my work off of you and getting feedback, I'd appreciate it.

The manager of the local National Video Store sells videocassette recorders at discount prices. If the store does not have a video recorder in stock when a customer wants to buy one, it will lose the sale because the customer will purchase a recorder from one of the many local competitors. The problem is that the cost of renting warehouse space to keep enough recorders in inventory to meet all demand is excessively high. The manager has determined that if 85% of customer demand for recorders can be met, then the combined cost of lost sales and inventory will be minimized. The manager has estimated that monthly demand for recorders is normally distributed, with a mean of 175 recorders and a standard deviation of 55. Determine the number of recorders the manager should order each month to meet 85% of customer demand.

Okay, so my approach was to solve for "x", since I thought I knew Z, mean and standard deviation.

I came up with x = 221.75, or 222

But the professor made a comment on the post to use Z distribution; for .85 that would be .3023 and I don't know how that would fit into the equation of figuring out how many recorders to order.

Re: Have the mean, standard deviation and probability...but something seems off!

But the value of z is not given. You need to determine it. Once you have the correct value of z, you can calculate the number of recorders x to which it corresponds from z = (x – μ)/σ.

To determine the required value of z, you must make use of the knowledge that the manager must cover the lower 85% of monthly sales volumes that are normally distributed, in conjunction with a normal N(0,1) table or calculator.

The answer you arrived at is incorrect, as is the value 0.3023 you cited. Hint: The normal table usually gives the area under the N(0,1) curve starting at z = 0, and the normal curve is symmetric about z = 0.