Help With Statistics Normal Distribution.

Maths with Statistics help - normal distribution?

The statement P(79-a≤X≤79+b) = 0.6463, help is the simple way of asking this question. The best way to ask this question is:

Let X be a normal random variable with mean 79 and variance 144.


Find a and b such that a symmetric interval about the mean exists with the following: P(79-a≤X≤79+b) = 0.646


For any normal random variable X with mean μ and standard deviation σ , X ~ Normal( μ , σ ), (note that in most textbooks and literature the notation is with the variance, i.e., X ~ Normal( μ , σ² ). Most software denotes the normal with just the standard deviation.)

You can translate into standard normal units by:
Z = ( X - μ ) / σ

Moving from the standard normal back to the original distribuiton using:
X = μ + Z * σ

Where Z ~ Normal( μ = 0, σ = 1). You can then use the standard normal cdf tables to get probabilities.

If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central Limit Theorem.

If a sample of size is is drawn from a population with mean μ and standard deviation σ then the sample average xBar is normally distributed

with mean μ and standard deviation σ /√(n)

An applet for finding the values
http://www-stat.stanford.edu/~naras/jsm/...

calculator
http://stattrek.com/Tables/normal.aspx

how to read the tables
http://rlbroderson.tripod.com/statistics...

In this question we have
X ~ Normal( μx = 79 , σx² = 144 )
X ~ Normal( μx = 79 , σx = 12 )





we start with finding the tail probabilities
the area in the tails is 1 - 0.6463 = 0.3537

P( X < 79 - a ) = 0.3537/2, i.e., half of the to total area in the tails.

P( X < 79 - a) = 0.17685

we know from the standard normal that:

P(Z < -0.9274364) = 0.17685

the equation for the transformation from X to Z is:

(a - 79) / 12 = -0.9274364
a = 67.87076


(b - 79) / 12 = 0.9274364
b = 90.12924


Here is the check

Find P( 67.87076 < X < 90.12924 )
= P( ( 67.87076 - 79 ) / 12 < ( X - μ ) / σ < ( 90.12924 - 79 ) / 12 )
= P( -0.9274367 < Z < 0.9274367 )
= P( Z < 0.9274367 ) - P( Z < -0.9274367 )
= 0.82315 - 0.1768499
= 0.6463001

Help with statistics normal distribution problem?

You need a normal distribution (z-score) table to complete these. The z-score is basically the proportion of standard deviations the measure represents.

a. An IQ of 100 is (100-110)/25 = -0.4 standard deviations (below the mean).
The area below the curve (from z = -0.4 to zero) is 0.1554 or 15.54%.
The area below the curve on the entire positive side is 50%.
Therefore, the percentage of those with an IQ above 100 is 15.54% + 50% = 65.54%.

b. The lowest 25% of the distribution is the z-score where approximately 25% of the area (on the negative side from the z-score to zero AND from the z-score and lower) is below the curve. This z-score is about -0.675.
-0.675 * 25 = -16.875
110 – 16.875 = 93.125
IQs of 93.125 and lower are in the lowest 25% of the distribution.

c. The highest 5% of the distribution is the z-score where approximately 45% of the area (on the positive side from the z-score to zero) is below the curve. This z-score is about 1.645.
1.645 * 25 = 41.125
110 + 41.125 = 151.125
IQs of 151.125 and higher are in the highest 5% of the distribution.

I'm confused on a statistics problem with normal distribution. help please?

Kamal has 30 hens. The probability that any hen lays an egg on any day is 0.7. Hens do not lay more
than one egg per day, and the days on which a hen lays an egg are independent.
(i) Calculate the probability that, on any particular day, Kamal’s hens lay exactly 24 eggs. [2]
(ii) Use a suitable approximation to calculate the probability that Kamal’s hens lay fewer than 20 eggs
on any particular day. [5]

i got the first part fine...but when i do the second part, i keep getting negative numbers:
it's mean = 115, standard deviation = unknown
P(X>103)=0.8
P(Z>(103-115)/S.D.))=0.8
P(Z>(-12/S.D.))=.8

the -12 messes me up! can someone tell me how to get rid of this, because i'm not sure how, but i know there is a way :(
thanxxx :)

Statistics and normal distribution question, I need help with calculating z-scores, etc!!?

Here is the question:

The average length of a hospital stay is 5.9 days. If we assume a normal distribution and a standard deviation of 1.7 days, 15% of hospital stays are less than how many days? Twenty-five percent of hospital stays are longer than how many days?

I am trying to do this solely on the TI-83 calculator, which is what my teacher is requiring of us. I am not using the table that you can use to find area and what not, just calculator. I desperately need someone to explain how to get the z-score from the area percentage, and the answer they got!!! Thanks a lot in advance!!!

Kyle

Help with maths STATISTICS- NORMAL DISTRIBUTION?

Q1) The heights of an adult female population are normally distributed with mean 162 cm and s.d 7.5 cm

b) Sarah is a young girl. She is told that she is at the 60th percentile for height .

Assuming that she remains at the 60th percentile, estimate the height as an adult.
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Can someone take me through the steps please, I find this question quite tricky and haven't been able to solve it yet... ''/

Statistics:Normal Distribution Question; Need help getting started!?

Question 1.
The U.S. Marine Corps requires that men have heights between 64 in. and 78 in. (The National Health Survey shows that heights of men are normally distributed with a mean of 69.0 in. and a standard deviation of 2.8 in.)

1. Find the percentage of men meeting those height requirements. Are too many men denied the opportunity to join the Marines because they are too short or too tall?

2. If you are appointed to be the Secretary of Defense and you want to change the requirement so that only the shortest 2% and tallest 2% of all men are rejected, what are the new minimum and maximum height requirements?

3. If 64 men are randomly selected, find the probability that their mean height is greater than 68.0 in.