Please help me understand this is asking and how to solve.
P(-2>Z and Z<-1)=
and
X~N(3,4). Find value c such that P(c<x)=.7
Thank You!
P(-2>Z and Z<-1)
or more commonly written as
P(-2 > Z <-1)
wants you to find the probability that the z-score of the normal binomial distribution lies between
-2 and -1
X~N(3,4) or X~N(3,2^2)
means X is normally distributed with a mean of 3 and a standard deviation of 2
These are common notations you should find in your text or in your class notes.
You must have been supplied with tables of the normal distribution.
If not you will find this applet extremely useful.
http://davidmlane.com/normal.html
What is the probability that the average lifetime of 25 bulbs would be no more than 1550 given mean of 1500 and standard deviation of 200 hours.
I know that it is asking P(x less than or equal to 1550), x is 25, u equal 1500 and sd is 200. I'm having problems getting to the final answer.
Thank YOu!
I will assume that you are working from "tables", so we need the z-score for 1550
z-score = (1550 - 1500)/200 = .25
look at
http://math.arizona.edu/~rsims/ma464/standardnormaltable.pdf
and find .25
(2nd page , find .2 in 1st column, then under .05)
to find .59871
which is the probability of your event.
The beauty of
http://davidmlane.com/normal.html
is that we don't even have to bother with z-scores.
just enter the mean of 1500
the sd of 200
click on the "below" button after entering 1550
to get .5987
To understand and solve the first question, let's break it down step by step:
P(-2 > Z and Z < -1)
Here, Z represents a standard normal random variable, which follows a standard normal distribution with mean 0 and standard deviation 1.
To solve this probability question, we need to find the probability that Z is between -2 and -1.
1. Start by sketching the standard normal distribution on a graph. The distribution is symmetric about the mean (0) and has values ranging from negative to positive infinity.
2. The probability we need to find, P(-2 > Z and Z < -1), corresponds to the shaded area under the curve between -2 and -1.
3. To calculate this probability, we can use a table of standard normal probabilities or a calculator. The table or calculator will provide the probability associated with the Z-scores of -2 and -1.
4. Alternatively, if you're using a standard normal distribution table, you can find the probability of Z being less than -1 and subtract the probability of Z being less than -2. This is because the probabilities in the table usually represent the area to the left of a given Z-score.
5. After finding the individual probabilities, subtract the probability of Z being less than -2 from the probability of Z being less than -1 to get the final probability.
Moving on to the second question:
X ~ N(3, 4)
This notation means that X is a random variable following a normal distribution with mean 3 and standard deviation 4.
To find the value c such that P(c < X) = 0.7:
1. Recognize that P(c < X) represents the probability that X is greater than c.
2. To find this value, you can use a standard normal distribution table or a calculator. However, since the question mentions a normal distribution with a specific mean and standard deviation, we need to standardize the variable X.
3. Standardizing means transforming the X-values into Z-scores, which have a mean of 0 and a standard deviation of 1. This helps us use the standard normal distribution table.
4. To standardize X, subtract the mean (3) from each X-value and then divide by the standard deviation (4). This will convert X into Z.
5. Once you have the standardized value Z, use a standard normal distribution table or calculator to find the Z-score corresponding to a probability of 0.7. This will give you the value of c, which you can then transform back to the X-scale by multiplying by the standard deviation and adding the mean.
Remember, when using a standard normal distribution table, look up the area in the table that corresponds to the probability given in the question. This will provide the corresponding Z-score, which can then be transformed back to the original variable scale if needed.