(1 point) Women's weights are normally distributed with a mean given by μ=143 lb and a standard deviation given by σ=29 lb. Find the fifth decile, D5, which separates the bottom 50% from the top 50%
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To find the fifth decile, which separates the bottom 50% from the top 50%, we need to find the value of weight that corresponds to this position in the normal distribution.
Here's how you can do it step-by-step:
Step 1: Calculate the z-score
The z-score represents the number of standard deviations away from the mean a specific value is. It is calculated using the formula:
z = (x - μ) / σ
In this case, we want to find the weight that separates the bottom 50% from the top 50%, which is equivalent to finding the z-score that corresponds to the 50th percentile. Since the normal distribution is symmetrical, this will be a z-score of 0.
Step 2: Convert the z-score to the corresponding weight value
After calculating the z-score, we need to convert it back to a weight value using the formula:
x = μ + z * σ
In this case, since the z-score is 0, the formula simplifies to:
x = μ
Where:
x = weight value corresponding to the fifth decile
μ = mean weight (given as 143 lb)
σ = standard deviation (given as 29 lb)
Step 3: Calculate the fifth decile weight
Substituting the given values into the formula:
x = 143 lb
Therefore, the weight corresponding to the fifth decile is 143 lb.