Probability Homework Please Help

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kevsdoll

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I am so unsure of this problem.
Suppose that the wrapper of a candy bar lists its weight as 8oz.The actual weights of candy bars vary to some extent. Suppose that these actual weights vary according to a normal distribution with a mean of 8.5 oz and a standard deviation of .325 oz.
a) What proportion of the candy bars weigh less than the advertised weight of 8 oz?
b) If the manufacturer wants to decrease this proportion by changing the mean, should it increase or decrease the mean? Explain without performing calculations.
b) If the manufacturer wants to decrease this proportion by changing the standard deviation, should it increase or decrease the standard deviation? Explain without performing calculations.
mean = 8.5 standard deviation = .325 My first thought is to get the z score. where z = 8-8.5/.325 = -1.54 then looking the value up on the standard normal distributions table. The value on the table is .0618. But this gives me the probability of the candy bars being under the mean of 8.5 oz. Then I thought I would have to use the empirical rule . Being approximatly 1.5 standard deviations below the mean. I am so confused . Please help.
 
a)

z = (8 - 8.5)/0.325

This should ring some bells.
 
I have that in the original message. I need to know how to find the percentage.
 
kevsdoll said:
I have that in the original message. I need to know how to find the percentage.
Click to expand...
WHAT percentage?

"The value on the table is .0618"
That means 6.18%.
Is that what you're asking?
 
kevsdoll said:
The value [from] the table is .0618. But this gives me the probability of the candy bars being under the mean of 8.5 oz.
Click to expand...


I don't think that the value 0.0618 relates to bars weighing less than the mean. The mean of 8.5 oz is the average weight (based on some sampling). So, we expect that roughly half (50%) of the bars produced weigh less than 8.5 oz.

The z-score that you calculated relates to bars that are less than 8 oz, not less than the mean, because (in the formula for z) you calculated -1.54 from using the value 8.

The total area under the normal curve represents 100% of a population. Z-score tables give a relative percent of this total area; specifically, the area under the curve to the LEFT of the z-score. So, 0.0618 represents the proportion of bars that weigh less than 8 oz.

In other words, I'm thinking that, if the probability of getting a bar less than 8 oz is 0.0618, then that corresponds (statistically) to 6.18% of all bars weighing less than 8 oz.

(Did I say these things correctly? My knowledge of statistcs is "rusty".)
 
Part of success in statistics is knowing when you are done.

I overlooked that you were struggling with this. Good work on the arithmetic and the table usage. Feel free to get up to speed on the knowing you are done part.
 
Thank yo guys so much. I did not think to change the z score to the percentage. I was looking to deep into the problem.
 
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