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Let’s say the random variable X is normally distributed with a mean (µ) = 35 and standard deviation (σ) = 5. Can you find P(25 < X < 45), i.e., the probability of the variable falling between 25 and 45? Note 1: Report the answer as a number rounded off to two digits after the decimal point.Note 2: You’re not expected to know the right answer at this. This question is given just to get you thinking in the right direction

Question

Let’s say the random variable X is normally distributed with a mean (µ) = 35 and standard deviation (σ) = 5. Can you find P(25 < X < 45), i.e., the probability of the variable falling between 25 and 45? Note 1: Report the answer as a number rounded off to two digits after the decimal point.Note 2: You’re not expected to know the right answer at this. This question is given just to get you thinking in the right direction

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Solution

To find the probability of the variable falling between 25 and 45, we first need to convert these values to z-scores. The z-score is a measure of how many standard deviations an element is from the mean.

The formula to calculate the z-score is:

Z = (X - µ) / σ

Where: X is the value from the dataset, µ is the mean of the dataset, and σ is the standard deviation of the dataset.

First, let's calculate the z-score for 25:

Z_25 = (25 - 35) / 5 = -2

Next, let's calculate the z-score for 45:

Z_45 = (45 - 35) / 5 = 2

Now, we need to find the probability that the z-score is between -2 and 2. This is the same as finding the area under the standard normal curve between -2 and 2.

Looking up these values in the z-table, we find that:

P(Z < 2) = 0.9772 (this is the area to the left of Z = 2) P(Z < -2) = 0.0228 (this is the area to the left of Z = -2)

To find the probability that Z is between -2 and 2, we subtract the smaller probability from the larger one:

P(-2 < Z < 2) = P(Z < 2) - P(Z < -2) = 0.9772 - 0.0228 = 0.9544

So, the probability of the variable falling between 25 and 45 is approximately 0.95, or 95%, when rounded off to two digits after the decimal point.

This problem has been solved

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