Respuesta :
Answer:
About 34% of the weights are between 3280 grams and 3720 grams.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 3500 grams, standard deviation of 500 grams.
This means that [tex]\mu = 3500, \sigma = 500[/tex]
About 34% of the weights are between what two values?
Between the 50 - (34/2) = 33rd percentile and the 50 + (34/2) = 67th percentile.
33rd percentile:
X when Z has a p-value of 0.33, so X when Z = -0.44.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.44 = \frac{X - 3500}{500}[/tex]
[tex]X - 3500 = -0.44*500[/tex]
[tex]X = 3280[/tex]
67th percentile:
X when Z has a p-value of 0.67, so X when Z = 0.44.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.44 = \frac{X - 3500}{500}[/tex]
[tex]X - 3500 = 0.44*500[/tex]
[tex]X = 3720[/tex]
About 34% of the weights are between 3280 grams and 3720 grams.
