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I'm taking a quiz and the only thing it's giving me is "Given a standard normal distrubution find the percent of data that is to the left of 2.53." I've looked in my book and still can't find anything. Can anyone help?

2007-06-29 15:53:24 · 8 answers · asked by gods_sold_out_servent 2 in Science & Mathematics Mathematics

8 answers

.9943
There is a table of normal probabilities in the back of your book.
You need to understand how to do various problems that use this table.

2007-06-29 16:01:57 · answer #1 · answered by fcas80 7 · 1 0

You need the mean and standard deviation. Subtract mean and your 2.53. Divide what you get by the standard deviation. What you get is the z score. There's a table to convert z score into % of data to the left of that.

If you aren't given mean and σ, then assume mean = 0 and σ = 1. In that case, 2.53 is your z and percent of data left of that is 99.4297%.

2007-06-29 16:12:37 · answer #2 · answered by Philo 7 · 0 0

I don't know what subject that is in math. I think that the most advanced math I can do is geometry. For example, the formula for finding the area of a circle is Pi r squared. I learned this last year in fifth grade from the best math teacher ever! Mrs. Sauli. It doesn't hurt that she nice too.

2007-06-29 15:58:29 · answer #3 · answered by Peanut to the rescue! 4 · 0 0

Yep: pay attention in class, and you will understand. Apparently you didn't, and now you're expecting total strangers to bail your *** out. Bad strategy, man: you don't learn anything that way.

2007-06-29 15:56:51 · answer #4 · answered by poorcocoboiboi 6 · 2 0

I don't think you should get help if you are taking a quiz! Good luck though :)

2007-06-29 16:04:17 · answer #5 · answered by triplea 3 · 0 0

tou should asked a tutor

2007-06-29 18:59:37 · answer #6 · answered by tinel 1 · 0 0

pay attention

listen carefully and attentively

no daydreaming

2007-06-29 16:00:05 · answer #7 · answered by Maxine 3 · 0 0

you need to know the mean in order to answer this.

2007-06-29 15:56:57 · answer #8 · answered by Anonymous · 1 0

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