Standard Normal Distribution

telepthy

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The number of calories eaten daily by a person is normally distributed with a mean of 3250 calories and a standard deviation of 550 calories.


a) What is the probability that a person eats less than 1963 calories daily?

b)What is the probability that a person eats more than 5345.5 calories daily?

c) What is the probability that a person eats between 3794.5 calories and 4064 calories daily?

d) How many calories must a person eat daily to be in the lowest 20%?
 
a) What do you think you need to do first?
 
Yes, your result is correct. I would have used:

\(\displaystyle P(X)=1-(0.5+0.4904)=0.0096\)

But, I have an old-school table that gives the area between the mean and the z-score.

For part b) it is essentially the same method, only on the opposite side of the mean...
 
(5345.5-3250)/550
3.81 z-score
(0.5+.4993)= .9993
1-.9993
.0007
did I do something wrong?

Can you elaborate how to do c & d?
 
My chart gives 0.4999, and using numeric integration I find it is closer to 0.499930516604, so I would use:

\(\displaystyle P(X)=1-(0.5+0.4999)=0.0001\)

For part c), you want to find the area between the mean and the larger value, and subtract from this the area between the mean and the smaller value, and this will give you the area between the smaller value and the larger value. This works because both values are greater than the mean. If this is not clear, I recommend drawing a sketch, and this should make it clear.

What would you do if the two values were on opposite sides of the mean?
 
c)

(3794.5-1963)/550
3.33= .445

(4064-1963)/550
3.82=.499

.445-499= -0.054

My chart stops at 3.59. Maybe i'm getting the z-score wrong?
 
You are using 1963 where you should be using the mean of 3250 in your standardization of the data (converting the data to z-scores).
 
(3794.5-3250)/550
.99= .3389

(4064-3250)/550
1.48= .4306

.3389-.4306= -.0917
 
You have the correct z-scores and areas, you just want to subtract the smaller from the larger to get 0.0917.

Now for part d) you want to find the x value for which 80% of the data is to the right of it. Do you have any ideas how you might accomplish this?
 
Do I add .5 to .8 or .2, then find whatever's closest to it from the chart to get the z-score, then subtract from the mean?
 
We know 0.5 will be to the right of the mean, so we want to find what z-score will give us the addition 0.3 to the left of the mean, but to the right of the x value. So, look on your chart to find which z-score is closest to an area of 0.3, then convert that z-score into an x value, keeping in mind is is to the left of the mean, which means you want to attach a negative sign to the z-score. Then, find the x-value associated with that z-score:

Since we have:

\(\displaystyle z=\dfrac{x-\mu}{\sigma}\)

you want to solve this for \(\displaystyle x\), then plug in all of the data.
 
You have the correct z-score, but you want to first solve the formula to convert an x value to a z-score for x so that you now have a formula to convert a z-score into an x value:

\(\displaystyle x=z\sigma+\mu\)
 
\(\displaystyle x=-0.84\cdot550+3250=?\)
 
Yes...do you see now what I did? I only ask because I want you to have a firm grasp of what happened, and I will be glad to offer clarification if anything is unclear. :mrgreen:
 
Glad to help...I am always happy to help someone who is willing to stick it out and work the problem to its resolution. :cool:

Welcome to FMH too, by the way! :mrgreen:
 
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