Question

If \(X\) is normally distributed with mean 1 and variance 4 , use the tables to find \(P\\{2

Short answer

The probability that a normally distributed random variable \(X\) with mean 1 and variance 4 falls between 2 and 3 is approximately 0.1498, or 14.98%.

Step-by-step solution

Identify the given parameters.

The problem states that \(X\) is normally distributed with a mean (\(\mu\)) of 1 and a variance (\(\sigma^2\)) of 4. From the given variance, we can find the standard deviation (\(\sigma\)) by taking the square root of the variance. Therefore, $$ \sigma = \sqrt{4} = 2. $$

Standardize the random variable using Z-scores.

To calculate the Z-scores, we need to subtract the mean from the given values and divide by the standard deviation. The Z-score formula is given by: $$ Z = \frac{X - \mu}{\sigma} $$ We will apply this formula to both the lower (\(X_1 = 2\)) and upper (\(X_2 = 3\)) boundaries of our desired range. Our goal is to find \(P\{Z_1 < Z < Z_2\}\), with \(Z_1\) and \(Z_2\) being the Z-scores corresponding to \(X_1\) and \(X_2\). Calculating these Z-scores, we have: $$ Z_1 = \frac{X_1 - \mu}{\sigma} = \frac{2 - 1}{2} = 0.5 $$ $$ Z_2 = \frac{X_2 - \mu}{\sigma} = \frac{3 - 1}{2} = 1 $$

Find the probabilities in the Z-table.

Now, using a standard normal table (Z-table), we can lookup the probabilities corresponding to the calculated Z-scores \(Z_1 = 0.5\) and \(Z_2 = 1\). These probabilities represent the area under the curve to the left of each Z-score. According to the Z-table: \(P\{Z < 0.5 \} = 0.6915\) and \(P\{ Z < 1\} = 0.8413\)

Calculate the probability for the desired range.

To find the probability \(P\{2<X<3\}\), we can subtract the probability of \(Z_1\) from the probability of \(Z_2\). Using the probabilities we found in the Z-table: $$ P\{2 < X < 3\} = P\{0.5 < Z < 1\} = P\{Z < 1\} - P\{Z < 0.5\} = 0.8413 - 0.6915 = 0.1498 $$ So, the probability of the random variable \(X\) falling between 2 and 3 is approximately 0.1498, or 14.98%.