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The weight, in pounds, of a certain type of adult squirrel is normally distributed with a mean of 4.1 pounds and a standard deviation of 0.5 pounds. What percentage of squirrels have a weight between 3 and 5 pounds?

Answers

Answer 1

Answer:

Answer:

95.02% of squirrels have a weight between 3 and 5 pounds.

Step-by-step explanation:

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.1 pounds

Standard Deviation, σ = 0.5 pounds

We are given that the distribution of weight is a bell shaped distribution that is a normal distribution.

Formula:

$z_(score) = \displaystyle(x-\mu)/(\sigma)$

P(weight between 3 and 5 pounds)

$P(3 \leq x \leq 5) = P(\displaystyle(3 - 4.1)/(0.5) \leq z \leq \displaystyle(5-4.1)/(0.5)) = P(-2.2 \leq z \leq 1.8)\n\n= P(z \leq 1.8) - P(z < -2.2)\n= 0.9641- 0.0139 = 0.9502= 95.02\%$

$P(3 \leq x \leq 5) = 95.02\%$

95.02% of squirrels have a weight between 3 and 5 pounds.

Answer 2

Answer:

Final answer:

Using the concepts of normal distribution and Z-scores, it is found that approximately 95% of the squirrels have weights between 3 and 5 pounds.

Explanation:

The problem involves the use of the concept of normal distribution in statistics. In this case, we have a mean of 4.1 pounds and a standard deviation of 0.5 pounds. We have two weight points, namely 3 pounds and 5 pounds. To find out the percentage in this interval, we need to convert the weights into z-scores, which standardizes them. The formula for this is Z = (X - μ) / σ, where X is the weight point, μ is the mean, and σ is the standard deviation.

The Z-score for 3 pounds is Z1 = (3 - 4.1) / 0.5 = -2.2, and the Z-score for 5 pounds is Z2 = (5 - 4.1) / 0.5 = 1.8. Looking these Z-scores up in the standard normal table (or using a calculator with a normal distribution function), we get 0.9857 for Z1 and 0.9641 for Z2. The difference between these values gives the percentage of squirrels within the interval, which is 0.9641 - (1 - 0.9857) = 0.950 or 95%. Hence, approximately 95% of the squirrels have a weight between 3 and 5 pounds.

Learn more about Normal Distribution here:

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Answers

Answer:

The original length of the rope is 140

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Judging on the basis of experience, a politician claims that 57% of voters in a certain area have voted for an independent candidate in past elections. Suppose you surveyed 25 randomly selected people in that area, and 18 of them reported having voted for an independent candidate. The null hypothesis is that the overall proportion of voters in the area that have voted for an independent candidate is 57%. What value of the test statistic should you report?

Answers

Answer: z= 1.51

Step-by-step explanation:

Test statistic for proportion is given by :-

$z=\frac{p-P}{\sqrt{(PQ)/(n)}}$

Where n is sample size ,p is the sample proportion , P Is the population proportion and Q =1 - P.

Given : P=57% = 0.57

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Test statistic for proportion will be :-

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What is the next number in the sequence. 1,121,12321, 1234321

The next number in the sequence is _____

Answers

Answer:

123454321

Step-by-step explanation:

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A cylindrical package to be sent by a postal service can have a maximum combined length and girth (perimeter of a cross section) of 141 inches. Find the dimensions of the package of maximum volume that can be sent. (The cross section is circular.)

Answers

Answer:

Length = 47 in

Radius = 47/π in

Step-by-step explanation:

Let 'h' be the length of the package, and 'r' be the radius of the cross section.

The length and girth combined are:

$L+G=141=h+2\pi r\nh=141-2\pi r$

The volume of the cylindrical package is:

$V=A_b*h\nV=\pi r^2*h$

Rewriting the volume as a function of 'r':

$V=\pi r^2*h\nV=\pi r^2*(141-2\pi r)\nV=141\pi r^2-2\pi^2 r^3$

The value of 'r' for which the derivate of the volume function is zero yields the maximum volume:

$V=141\pi r^2-2\pi^2 r^3\n(dV)/(dr)=282\pi r-6\pi^2r^2=0\n 6\pi r=282\nr=(47)/(\pi) \ in$

The length is:

$h=141-2\pi r=141-2\pi*(47)/(\pi)\nh=47\ in$

The dimensions that yield the maximum volume are:

Length = 47 in

Radius = 47/π in

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Answers

Answer:

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Answers

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