The weights of a certain dog breed are approximately normally distributed with a mean of 49 pounds, and a standard deviation of 6 pounds. Use your graphing calculator to answer the following questions. Write your answers in percent form. Required:
a. Find the 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.
c. Find the percentage of dogs of this breed that weigh more than 49 pounds.

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 and standard deviation , the z-score of a measure X is given by:

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

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

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:

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:

has a p-value of 0.5.

1 - 0.5 = 0.5 = 50% of dogs of this breed that weigh more than 49 pounds.

687.09 dollars       happy now

step-by-step explanation: 6870

1857 times .63 equals 1169.91 subtract that from 1857 you get 687.09

yes because there is exactly one y value for every x value

for function there can't be more than one value for every x value