Question 1168193: The birth weights for twins are normally distributed with a mean of 2350 grams and a standard deviation of 645 grams. Identify z-scores and determine the birth weight of 2255 could be considered unusual.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Solution:
The birth weights for twins are normally distributed with a mean ($\mu$) of 2350 grams and a standard deviation ($\sigma$) of 645 grams. We want to determine if a birth weight of 2255 grams could be considered unusual.
To do this, we first calculate the z-score for a birth weight of 2255 grams using the formula:
$z = \frac{x - \mu}{\sigma}$
where $x$ is the birth weight, $\mu$ is the mean birth weight, and $\sigma$ is the standard deviation.
Given:
$x = 2255$ grams
$\mu = 2350$ grams
$\sigma = 645$ grams
Calculate the z-score:
$z = \frac{2255 - 2350}{645} = \frac{-95}{645} \approx -0.147$
Rounding the z-score to two decimal places, we get $z \approx -0.15$.
Now, we need to determine if this z-score indicates an unusual birth weight. A common rule of thumb is that a data point is considered unusual if its z-score is less than -2 or greater than 2. This corresponds to data points that are more than 2 standard deviations away from the mean.
In this case, the z-score of -0.15 is between -2 and 2. Therefore, a birth weight of 2255 grams is not considered unusual based on this criterion. It falls within the typical range of birth weights for twins according to the given distribution.
Final Answer: The final answer is $\boxed{z \approx -0.15, \text{ not unusual}}$
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