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Find the z-scores that separate the middle 52% of the distribution from the area in the tail of the standard normal distribution.

User Hgtcl
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Final Answer:

The z-scores that separate the middle 52% of the distribution from the area in the tail of the standard normal distribution are approximately -0.848 and 0.848.

Step-by-step explanation:

The standard normal distribution has a total of 100% probability density, with 50% on either side of the mean (z-score of 0). To find the z-scores that encompass the middle 52%, first, determine the remaining percentage in the tails, which is (100% - 52%) / 2 = 24%.

Consulting the standard normal distribution table or using a calculator, locate the z-scores corresponding to 24% in one tail. This value is approximately 0.848. As the distribution is symmetrical, the z-score for the lower tail will be the negative of this value, approximately -0.848.

These z-scores represent the boundary where the middle 52% of the distribution lies, leaving 24% in each tail beyond these values. Utilizing these z-scores in standard normal calculations allows for identification of values falling within this specific range of the distribution.

User Navderm
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The z-scores that separate the middle 52% of the distribution from the area in the tail of the standard normal distribution are approximately Z1 ≈ -0.7112 and Z2 ≈ 0.6745.

To find the z-scores that separate the middle 52% of the distribution from the area in the tail of the standard normal distribution, you can follow these steps:

1. Understand the problem:

You want to find two z-scores, let's call them Z1 and Z2, such that the middle 52% of the standard normal distribution is between these two z-scores, and the remaining area (tail) is outside them.

2. Find the cumulative probabilities:

To do this, you need to find the cumulative probabilities associated with the middle 52% and the tail areas.

a) Middle 52%:

To find the middle 52%, you can start by finding the cumulative probability for the lower percentile. Since the middle 52% is symmetric around the mean (z = 0), you need to find the cumulative probability for the lower 24% (i.e., (100% - 52%) / 2 = 24%).

You can use a standard normal distribution table or a calculator to find this value. In many cases, you'll find the value for the 24th percentile to be approximately 0.2438.

b) Tail area:

The remaining area in the tails is 100% - 52% = 48%. Since it's evenly split between the two tails, each tail has an area of 24%.

3. Find the z-scores:

Now that you have the cumulative probability for the lower 24% (0.2438) and the tail area of 24%, you can use the z-table or a calculator to find the z-scores associated with these probabilities.

a) For the lower 24%, you can look up the z-score corresponding to the cumulative probability of 0.2438. This will give you Z1.

b) For the upper 24% (tail area), you can look up the z-score corresponding to the cumulative probability of 1 - 0.24 = 0.76. This will give you Z2.

4. Calculate Z1 and Z2:

Using a standard normal distribution table or calculator, you can find the z-scores for the probabilities you found:

a) Z1 ≈ -0.7112 (for the lower 24%)

b) Z2 ≈ 0.6745 (for the upper 24%)

User Stellar Roki
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