Answer:
0.0344 = 3.44% probability that a newborn pup has a weight above 24.0 lbs.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 22.0 lbs and standard deviation of 1.1 lbs.
This means that ![\mu = 22, \sigma = 1.1](https://tex.z-dn.net/?f=%5Cmu%20%3D%2022%2C%20%5Csigma%20%3D%201.1)
Find the probability that a newborn pup has a weight above 24.0 lbs.
This is 1 subtracted by the p-value of Z when X = 24. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{24 - 22}{1.1}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B24%20-%2022%7D%7B1.1%7D)
![Z = 1.82](https://tex.z-dn.net/?f=Z%20%3D%201.82)
has a p-value of 0.9656.
1 - 0.9656 = 0.0344
0.0344 = 3.44% probability that a newborn pup has a weight above 24.0 lbs.