What is the probability of a z score being either less than -1.65 or greater than 1.65?
Understand the Problem
The question is asking for the probability associated with a standard normal distribution concerning a z score of -1.65 and 1.65. We need to calculate the area under the curve in both tails of the distribution for these z scores.
Answer
The total probability in both tails for z-scores of -1.65 and 1.65 is $0.0990$.
Answer for screen readers
The total area under the curve in both tails for z-scores of -1.65 and 1.65 is $0.0990$.
Steps to Solve
- Identify the z-scores
We are given two z-scores: $z_1 = -1.65$ and $z_2 = 1.65$.
- Use the standard normal distribution table
To find the area under the curve for each z-score, we will refer to the standard normal distribution table (Z-table). This table provides the cumulative probability for z-scores.
- Find the area for $z_1 = -1.65$
Looking up $z_1 = -1.65$ in the Z-table, we find:
$$ P(Z < -1.65) = 0.0495 $$
This represents the area to the left of $z_1$.
- Find the area for $z_2 = 1.65$
Next, we look up $z_2 = 1.65$ in the Z-table:
$$ P(Z < 1.65) = 0.9505 $$
This area represents the area to the left of $z_2$.
- Calculate the area in both tails
To find the area in both tails under the standard normal curve, we calculate the combined area for $z_1$ and the tail area for $z_2$. The area to the right of $z_2$ is:
$$ P(Z > 1.65) = 1 - P(Z < 1.65) = 1 - 0.9505 = 0.0495 $$
- Add both tail areas
Now, we can combine both areas:
$$ \text{Total Area} = P(Z < -1.65) + P(Z > 1.65) = 0.0495 + 0.0495 = 0.0990 $$
The total area under the curve in both tails for z-scores of -1.65 and 1.65 is $0.0990$.
More Information
This area of $0.0990$ represents the probability of obtaining a value in the tails of the distribution when dealing with a standard normal distribution. This can be useful in hypothesis testing and confidence interval calculations.
Tips
- Forgetting to subtract the cumulative probability from 1 for the right tail of the z-score.
- Misreading the Z-table, which can lead to incorrect cumulative probabilities.
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