Respuesta :
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:
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.
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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.
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- 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.
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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.
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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.
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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.
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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.
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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.
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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
