The weights of healthy adult male Labrador Retrievers are approximately Normally distributed with a mean of 77 pounds and a standard deviation of 6 pounds

A)Make an accurate sketch of the weights of Labrador Retrievers with the horizontal axis marked

B)What weight would represent the 25th percentile?

C)What weight would represent the 75th percentile?

D) What is the interquartile range of male Labrador weights?

E)What proportion of male adult Labrador retrievers weigh above 85 pounds?

F) What proportion of male adult Labrador retrievers weigh between 60 and 80 pounds?

G) What proportion of male adult Labrador retrievers weigh less than 65 pounds?

Answers

Answer 1

Answer:the photo is for part A.

B) the weight would be Invnorm(.25,77,6)=72.95

C)the weight would be 81.04 but this is the work invnorm(.75,77,6)=81.04

D)IQR=Q3-Q1

81.04-72.95=8.09

E)9.12 is the answer the work normalcdf(85,1000,77,6)=.0912

F).6892 or 68.92% are the answers the work normalcdf(60,80,77,6)=.6892

G).0227 or 2.27% is the answer and the work normalcdf(0,65,77,6)=.0227

Step-by-step explanation:

The Weights Of Healthy Adult Male Labrador Retrievers Are Approximately Normally Distributed With A Mean
Answer 2

Using the normal distribution, we have that:

a) The sketch is given at the end.b) A weight of 72.95 pounds would represent the 25th percentile.c) A weight of 81.05 pounds would represent the 75th percentile.d) The interquartile range is of 8.1 pounds.e) 0.0918 = 9.18% of male adult Labrador retrievers weigh above 85 pounds.f) 0.6892 = 68.92% of male adult Labrador retrievers weigh between 60 and 80 pounds.g) 0.0228 = 2.28% of male adult Labrador retrievers weigh less than 65 pounds.

----------------------

Problems of normal distribution 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 77 means that [tex]\mu = 77[/tex]Standard deviation of 6 means that [tex]\sigma = 6[/tex]As for item a, the sketch is given at the end of the question.

----------------------

Item b:

The weight is X when Z has a p-value of 0.25, so X when Z = -0.675.

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

[tex]-0.675 = \frac{X - 77}{6}[/tex]

[tex]X - 77 = -0.675(6)[/tex]

[tex]X = 72.95[/tex]

A weight of 72.95 pounds would represent the 25th percentile.

----------------------

Item c:

The weight is X when Z has a p-value of 0.75, so X when Z = 0.675.

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

[tex]0.675 = \frac{X - 77}{6}[/tex]

[tex]X - 77 = 0.675(6)[/tex]

[tex]X = 81.05[/tex]

A weight of 81.05 pounds would represent the 75th percentile.

----------------------

Item d:

The interquartile range is the difference between the 75th and the 25 percentile, thus:

[tex]81.05 - 72.95 = 8.1[/tex]

The interquartile range is of 8.1 pounds.

----------------------

Item e:

The proportion is 1 subtracted by the p-value of Z when X = 85, thus:

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

[tex]Z = \frac{85 - 77}{6}[/tex]

[tex]Z = 1.33[/tex]

[tex]Z = 1.33[/tex] has a p-value of 0.9082.

1 - 0.9082 = 0.0918.

0.0918 = 9.18% of male adult Labrador retrievers weigh above 85 pounds.

----------------------

Item f:

The proportion is the p-value of Z when X = 80 subtracted by the p-value of Z when X = 60, thus:

X = 80:

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

[tex]Z = \frac{80 - 77}{6}[/tex]

[tex]Z = 0.5[/tex]

[tex]Z = 0.5[/tex] has a p-value of 0.6915.

X = 60:

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

[tex]Z = \frac{60 - 77}{6}[/tex]

[tex]Z = -2.83[/tex]

[tex]Z = -2.83[/tex] has a p-value of 0.0023.

0.6915 - 0.0023 = 0.6892.

0.6892 = 68.92% of male adult Labrador retrievers weigh between 60 and 80 pounds.

----------------------

Item g:

This proportion is the p-value of Z when X = 65, thus:

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

[tex]Z = \frac{65 - 77}{6}[/tex]

[tex]Z = -2[/tex]

[tex]Z = -2[/tex] has a p-value of 0.0228.

0.0228 = 2.28% of male adult Labrador retrievers weigh less than 65 pounds.

A similar problem is given at https://brainly.com/question/21628677

The Weights Of Healthy Adult Male Labrador Retrievers Are Approximately Normally Distributed With A Mean

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