Respuesta :
Answer:
a. 74.86%
b. 50%
c. 50%
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.
Normally distributed with a mean of 49 pounds, and a standard deviation of 6 pounds.
This means that [tex]\mu = 49, \sigma = 6[/tex]
a. Find the percentage of dogs of this breed that weigh less than 53 pounds.
The proportion is the p-value of Z when X = 53. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{53 - 49}{6}[/tex]
[tex]Z = 0.67[/tex]
[tex]Z = 0.67[/tex] has a p-value of 0.7486.
0.7486*100% = 74.86%, which is percentage of dogs of this breed that weigh less than 53 pounds.
b. Find the percentage of dogs of this breed that weigh less than 49 pounds.
p-value of Z when X = 49, so:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{49 - 49}{6}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] has a p-value of 0.5
0.5 = 50% of dogs of this breed that weigh less than 49 pounds.
c. Find the percentage of dogs of this breed that weigh more than 49 pounds.
1 subtracted by the p-value of Z when X = 49, so:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{49 - 49}{6}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] has a p-value of 0.5.
1 - 0.5 = 0.5 = 50% of dogs of this breed that weigh more than 49 pounds.
