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The ages of alligators in a swamp are normally distributed, with a mean of 50 years and a standard deviation of 4. Approximately what percentage of the alligators are between 45 and 55 years old?


78.87%

10.56%

89.44%

67.76%

Question 2(Multiple Choice Worth 3 points)

(05.03 LC)


Wait times at an orthodontist's office are typically 17 minutes, with a standard deviation of 2 minutes. What percentage of people should be seen by the doctor between 11 and 23 minutes for this to be considered a normal distribution?


17%

68%

95%

99.7%

Respuesta :

pedrocarratore

Answer:

1) 78.87%

2) 99.7%

Step-by-step explanation:

The z-score for a normal distribution is given by:

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

1- The given data is:

σ = 4 years

μ = 50 years

For X = 45 and X =55

[tex]z_1=\frac{45-50}{4}\\z_1=-1.25\\z_2=\frac{55-50}{4}\\z_2=1.25[/tex]

A z-score of -1.25 corresponds to the 10.57th percentile of a normal distribution, while a z-score of 1.25 is equivalent to the 89.44th percentile.

Therefore, the percentage of the alligators are between 45 and 55 years old is:

[tex]P=89.44-10.57=78.87\%[/tex]

78.87% of alligators.

2- The given data is:

σ = 2 minutes

μ = 17 minutes

For X = 11 and X =23

[tex]z_1=\frac{11-17}{2}\\z_1=-3\\z_2=\frac{23-17}{2}\\z_2=3[/tex]

A z-score of -3.0 corresponds to the 0.135th percentile of a normal distribution, while a z-score of 3.0 is equivalent to the 99.865th percentile.

Therefore, the percentage of people that should be seen by the doctor between 11 and 23 minutes is:

[tex]P=99.865-.0135=99.73\%[/tex]

99.7% of people.

Using the normal distribution, it is found that:

  • 78.87% of the alligators are between 45 and 55 years old.
  • 99.7% of people should be seen by the doctor between 11 and 23 minutes for this to be considered a normal distribution.

Normal Probability Distribution

In a normal distribution 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]

  • It 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, which is the percentile of X.
  • The Empirical Rule states that for a normal variable, 99.7% of the measures are within 3 standard deviations of the mean.

In this problem:

  • Mean of 50, hence [tex]\mu = 50[/tex].
  • Standard deviation of 4, hence [tex]\sigma = 4[/tex].

The proportion between 45 and 55 years old is the p-value of Z when X = 55 subtracted by the p-value of Z when X = 45, then:

X = 55:

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

[tex]Z = \frac{55 - 50}{4}[/tex]

[tex]Z = 1.25[/tex]

[tex]Z = 1.25[/tex] has a p-value of 0.8944.

X = 45:

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

[tex]Z = \frac{55 - 50}{4}[/tex]

[tex]Z = -1.25[/tex]

[tex]Z = -1.25[/tex] has a p-value of 0.1057.

0.8944 - 0.10567= 0.7887.

0.7887 = 78.87% of the alligators are between 45 and 55 years old.

For the second question, by the Empirical Rule, 99.7% of people should be seen by the doctor between 11 and 23 minutes for this to be considered a normal distribution.

To learn more about normal distribution, you can take a look at https://brainly.com/question/24663213

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