The weights of black bears are normally distributed with a mean of 264 kg and a standard deviation of 38 kg. Let X be the weight of a randomly selected black bear. Use this information for exercises 2 and 3.
2. If P(a < X < 264) = 0.1139, find the value of a. (You must show work and/or TI
calculator commands to receive credit. Round your answer to the nearest kg.)
3. If P(264 < X < b) = 0.2362, find the value of b. (You must show work and/or TI
calculator commands to receive credit. Round your answer to the nearest kg.)
you can play around with Z table stuff at
http://davidmlane.com/hyperstat/z_table.html
To find the value of 'a' in exercise 2, we need to calculate the z-score corresponding to the given probability and then convert it back to the original scale using the formula:
z = (X - mean) / standard deviation
First, we need to find the z-score using the cumulative distribution function (CDF) of the standard normal distribution. In this case, we know that P(a < X < 264) = 0.1139.
Using a standard normal distribution table or a TI calculator command (e.g., invNorm on TI-84), we can find the z-score corresponding to a cumulative probability of 0.1139.
In this case, the z-score is approximately -1.17.
Now, we can use the formula mentioned above to find the value of 'a':
-1.17 = (a - 264) / 38
Solving for 'a', we get:
a - 264 = -1.17 * 38
a - 264 = -44.46
a = 264 - 44.46
Therefore, a ≈ 219.54 kg (rounded to the nearest kg).
Moving on to exercise 3, we can follow a similar approach. We are asked to find the value of 'b' when P(264 < X < b) = 0.2362.
We can calculate the z-score corresponding to this probability using the standard normal distribution table or a TI calculator command. Let's say the z-score is denoted as 'z'.
Using the formula, we have:
0.2362 = (b - 264) / 38
Solving for 'b', we get:
b - 264 = 0.2362 * 38
b - 264 = 8.9896
b ≈ 264 + 8.9896
Therefore, b ≈ 272.9896 kg (rounded to the nearest kg).