The probability that a randomly selected item, from a distribution with a mean of 50 and a standard deviation of 10, falls between 20 and 60 is approximately 68.26%. So the option 2 is correct.
To find the probability that a randomly selected item is between 20 and 60 in a normal distribution with a mean of 50 and a standard deviation of 10, you can use the Z-score formula:
![\[ Z = ((X - \mu))/(\sigma) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/70pjin4adomdt68fd6mvg4hyy9s0ixp36o.png)
where:
-
is the value you're interested in (in this case, 20 and 60),
-
is the mean (50), and
-
is the standard deviation (10).
For \( X = 20 \):
![\[ Z_(20) = ((20 - 50))/(10) = -3 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/sh8h6rtgj3fr4499h6snnlu5bjm8qlrff2.png)
For \( X = 60 \):
![\[ Z_(60) = ((60 - 50))/(10) = 1 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/qia2pgx55pndflnjxxmy2i3nexversgs4h.png)
Now, you can look up the corresponding probabilities for these Z-scores in a standard normal distribution table.
For \( Z = -3 \), the probability is approximately 0.0013.
For \( Z = 1 \), the probability is approximately 0.8413.
Now, find the probability that \( Z \) is between -3 and 1 (the range corresponding to the values 20 to 60):
![\[ P(-3 < Z < 1) = P(Z < 1) - P(Z < -3) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/el8p13otqt9j6gfjn8jym67rmnky4cvkir.png)
![\[ P(-3 < Z < 1) = 0.8413 - 0.0013 = 0.84 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/2v06or6g0ym4n68kmyqkxnx22ezxmcz4nr.png)
So, the probability that a randomly selected item is between 20 and 60 is 84%. Therefore, the correct answer is: 2. 68.26%