The random variable X follows normal distribution.P(X<35)=0.2 and P(35<X<45)=0.65.Find mean and standard deviation.
Answer as fast as possible.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability and convert it to standard deviations.
For example: P < .2 is approximately -.84 SD.
Continue with your other data.
Again using
http://davidmlane.com/normal.html
with setting on "value from an area"
if P(X<35)=0.2 and P(35<X<45)=0.65 then P(x<45) = .85
entering .2 and clicking below gave me a z-score of -.841
(35-m)/s = -.841 ---> m - .841s = 35
entering .85 and clicking below gave me a z-score of 1.036
(45-m)/s = 1.036 ----> m + 1.036s = 45
subtract those two equations:
1.877s = 10
s = 5.3277
m = 39.48
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To find the mean and standard deviation of a random variable that follows a normal distribution, we need more information than just the probabilities given. We will need the mean and standard deviation of the random variable.
If we assume that the random variable X follows a standard normal distribution (mean = 0 and standard deviation = 1), we can use z-scores to find the mean and standard deviation. The z-score formula is given by:
z = (x - μ) / σ
where x is the observed value, μ is the mean, and σ is the standard deviation.
Using the z-score formula, we can estimate the mean and standard deviation for the given probabilities.
First, let's find the z-score for P(X < 35) = 0.2:
z = (35 - μ) / σ = -0.84
Next, let's find the z-scores for P(35 < X < 45) = 0.65. Since we don't have the exact values of μ and σ, we can't solve for them directly. But we can use the properties of the standard normal distribution to estimate the z-scores for the given probabilities.
Using a standard normal table or software, we can find the z-score that corresponds to a cumulative probability of 0.2. Looking up this value in the table, we find that the z-score is approximately -0.84.
Similarly, we can find the z-score for a cumulative probability of 0.65. Using the standard normal table or software, we find that the z-score is approximately 0.39.
Since the given probabilities are for the standard normal distribution, we can equate the z-scores to each other for z < 0.39. This allows us to see that the z-score of -0.84 corresponds to the value x = 35 and the z-score of 0.39 corresponds to the value x = 45. Therefore, in terms of the standard normal distribution, we can write:
P(-0.84 < Z < 0.39) = 0.65
Now, we need to find the values of μ and σ that satisfy this equation. Since we know that the mean of the standard normal distribution is 0, we can write:
0.65 = P(-0.84 < Z < 0.39) = P(35 < X < 45)
Since we assumed X follows a standard normal distribution, we can infer that the mean, μ, is 40 and the standard deviation, σ, is 5.
Therefore, for a random variable X that follows a normal distribution with μ = 40 and σ = 5, we have:
Mean (μ) = 40
Standard Deviation (σ) = 5