Chapter 6: Problem 26
Obtain the following probabilities for the standard normal distribution. a. \(P(z>-1.86)\) b. \(P(-.68 \leq z \leq 1.94)\) c. \(P(0 \leq z \leq 3.85)\) d. \(P(-4.34 \leq z \leq 0)\) e. \(P(z>4.82)\) f. \(P(z<-6.12)\)
Short Answer
Expert verified
The probabilities for each part of the problem are: a) 0.9686 or 96.86%, b) 0.7261 or 72.61%, c) 0.4999 or 49.99%, d) 0.5000 or 50.00%, e) approximately 0 or 0.00%, and f) approximately 0 or 0.00%.
Step by step solution
01
Obtain Probabilities for Part a
To find the probability \(P(z > -1.86)\), refer to the standard normal table or use a z-score calculator. This value represents the proportion of data to the right of z = -1.86 in the standard normal distribution. After looking up the value, the area to the left of z=-1.86 is 0.0314 or 3.14%, so the area to the right or \(P(z > -1.86)\) is 1-0.0314 = 0.9686 or 96.86%.
02
Obtain Probabilities for Part b
Now find the probability \(P(-.68 \leq z \leq 1.94)\). This value represents the proportion of data between z = -0.68 and z = 1.94 in the standard normal distribution. Firstly, find the area to the left of z=1.94, and the area to the left of z=-0.68, and then subtract one from the other. The probability is 0.9744 - 0.2483 = 0.7261 or 72.61%.
03
Obtain Probabilities for Part c
Next, find the probability \(P(0 \leq z \leq 3.85)\). This value represents the proportion of data between z = 0 and z = 3.85 in the standard normal distribution. The process is similar to that in step 2. The probability is 0.9999 - 0.5000 = 0.4999 or 49.99%.
04
Obtain Probabilities for Part d
Now find the probability \(P(-4.34 \leq z \leq 0)\). This value represents the proportion of data between z = -4.34 and z = 0 in the standard normal distribution. Subtract the area to the left of z=-4.34 from the area to the left of z=0. The probability is 0.5000 - 0.0000 = 0.5000 or 50.00%.
05
Obtain Probabilities for Part e
Find the probability \(P(z > 4.82)\). This value represents the proportion of data to the right of z = 4.82 in the standard normal distribution. The area to the right of this z-score is approximately 0 or 0.00% since it's an extreme value.
06
Obtain Probabilities for Part f
Finally, find the probability \(P(z < -6.12)\). The process is similar to step 5, and this value represents the proportion of data to the left of z = -6.12 in the standard normal distribution. The area to the left of this z-score is negligible and approximately 0 or 0.00% since it's a very extreme value.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Z-Score Probabilities
Z-scores are a statistical measurement that describes a value's relationship to the mean of a group of values. They are measured in terms of standard deviations from the mean. If a Z-score is 0, it indicates that the data point's score is identical to the mean score. A Z-score can be positive or negative, indicating whether the value is above or below the mean, respectively.
When dealing with Z-score probabilities, we are interested in determining the probability of a score occurring within our normal distribution. This is often represented as the area under the curve of our normal distribution for particular Z-scores. Knowing Z-score probabilities helps us understand how likely a data point is to occur relative to other data points.
Area Under the Curve
The area under the curve in a standard normal distribution is a crucial concept when it comes to probabilities. In statistics, when we aim to find the probability between two Z-scores, we calculate the area under the curve between these points.
This area represents the likelihood of a Z-score falling between the values you are interested in. The total area under the curve of a standard normal distribution is equal to one, which represents a probability of 100%.
Finding the area under the curve allows us to understand and visualize the likelihood of certain results occurring. For example, if the area under the curve from z = -0.68 to z = 1.94 is 0.7261, this means there's a 72.61% chance of a randomly selected score lying between these two values.
Standard Normal Table
The standard normal table, also referred to as the Z-table, is used to determine the area under the curve for any Z-score in a standard normal distribution. The Z-table shows the cumulative probability up to a particular Z-score, which is the area under the curve to the left of the Z-score.
To find the probability associated with a Z-score using the standard normal table:
- Find the corresponding Z-score value in the table. This will give you the cumulative probability.
- Subtracting the cumulative probability of a smaller Z-score from a larger one can get you the area between two scores.
- If you need the probability to the right of a Z-score, subtract the cumulative probability from 1.
Probability Calculation
Probability calculation in a standard normal distribution involves determining how likely it is for a particular event to occur. By using Z-scores, we can calculate these probabilities and make informed predictions.
Here is how you can perform probability calculations for different scenarios:
- For probabilities involving only one Z-score, use the area given directly in the standard normal table.
- For probabilities between two Z-scores, find the cumulative probabilities for both and subtract them to get the area in between.
- For probabilities to the right of a Z-score, subtract the cumulative probability from 1.
- For extremely low or high Z-scores (like e.g. 4.82 or -6.12), the probabilities are often so negligible that they are practically zero.