91Ó°ÊÓ

To understand normal distribution problems, one essential concept is the Z-score. The Z-score allows you to determine how many standard deviations an element is from the mean. This calculation helps us compare it to the overall data distribution.
The Z-score is calculated using the formula:
\[ Z = \frac{x - \mu}{\sigma} \]
In this formula:

• \( x \) is the value of interest (e.g., length of a specific pregnancy)
• \( \mu \) is the mean of the distribution
• \( \sigma \) is the standard deviation

In our example, the average length of pregnancies is 268 days, and the standard deviation is 15 days. For a pregnancy lasting 308 days, we find the Z-score to be approximately 2.67. This shows that 308 days is 2.67 standard deviations above the mean. Understanding Z-scores enables us to find probabilities associated with specific values in a normal distribution.

Once you have the Z-score, you can use it to find probabilities from the standard normal distribution (a normal distribution with a mean of 0 and a standard deviation of 1).
To find these probabilities, either use a Z-table or statistical software. The Z-table gives you the area under the normal curve to the left of the Z-score.
In our example, a Z-score of 2.67 corresponds to a cumulative probability of approximately 0.9963 or 99.63%.
To find the probability of an event happening beyond a certain Z-score, you subtract this value from 1:
\[ P(X \geq 308) = 1 - P(X < 308) = 1 - 0.9963 = 0.0037 \]
Converting this to percentage, there's a 0.37% probability that a pregnancy will last 308 days or longer. Such a low probability suggests that it is very rare.

To analyze any normal distribution, you must identify its parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)). These parameters summarize the central tendency and the spread of the distribution.
In our exercise, the mean length of pregnancies is \( \mu = 268 \) days, indicating that most pregnancies last around this time. The standard deviation \( \sigma = 15 \) days tells us that the lengths of pregnancies typically deviate from the mean by about 15 days.
These parameters are crucial because they allow us to use the normal distribution model to calculate probabilities and Z-scores. Without an accurate mean and standard deviation, our probability predictions would be inaccurate.

In medical statistics, determining thresholds such as the duration separating premature births is essential. A premature baby is typically one born before completing the full term. In our exercise, being premature means being in the lowest\(3 \%\) of pregnancy lengths.
First, we need to find the Z-score associated with the lowest \(3 \%\) of the distribution. From the Z-table, the Z-score corresponding to the bottom \(3 \%\) is roughly -1.88.
We use the Z-score formula to find the corresponding duration:
\[ Z = \frac{x - \mu}{\sigma} \]
Rearranging for \( x \):
\[ -1.88 = \frac{x - 268}{15} \]
Solving for \( x \):
\[ x = -1.88 \cdot 15 + 268 = -28.2 + 268 = 239.8 \approx 240 \]
Thus, pregnancies lasting 240 days or less are considered premature. This information helps doctors and hospitals plan for the special care that premature babies often require.