The weights of a certain dog breed are approximately normally distributed with a mean of 51 pounds, and a standard deviation of 5.8 pounds. Answer the following questions. Write your answers in percent form. Round your answers to the nearest tenth of a percent. a) Find the percentage of dogs of this breed that weigh less than 51 pounds. % b) Find the percentage of dogs of this breed that weigh less than 45 pounds. % c) Find the percentage of dogs of this breed that weigh more than 45 pounds. %

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pedrocarratore pedrocarratore · Answer 1 · 2019-12-06T06:56:49+01:00

Answer:

(a) P(X<51) = 50.0%

(b) P(X<45) = 15.1%

(c) P(X>45) = 84.9%

Step-by-step explanation:

Mean weight (μ) = 51 pounds

Standard deviation (σ) = 5.8 pounds

The z-score for any given weight, 'X', is given by the following expression;

[tex]z=\frac{X- \mu}{\sigma}[/tex]

a) P(X<51)

Since 51 pounds is exactly the mean weight, 50% of the dogs weight more than 51 pounds while 50% weight less than 51 pounds.

P(X<51) = 50.0%

b) P(X<45)

The z-score for X=45 is:

[tex]z=\frac{45- 51}{5.8}\\z=-1.034[/tex]

A z-score of -1.034 is equivalent to the 15.06-th percentile in a normal distribution. Therefore:

P(X<45) = 15.1%

c) P(X>45)

P(X>45) = 100% - P(X<45) = 100% - 15.1%

P(X>45) = 84.9%