In 1940 the average size of a U.S. farm was 174 acres. Let's say that the standard deviation was 51 acres. Suppose we randomly survey 45 farmers from 1940. I need help with the question below: The middle 50% of the distribution for x̅, the bounds of which form the distance represented by the IQR, lies between what two values?

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A: The provided information are:

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Q: The lengths of pregnancies in a small rural village are normally distributed with a mean of 265 days…

A: From the provided information, Mean (µ) = 265 Standard deviation (σ) = 15 X~N (265,15)

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Q: In 1940 the average size of a U.S. farm was 174 acres. Let's say that the standard deviation was 52…

A: It is given that the middle area of the distribution for X is 0.50.

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A: Here, mean lifespan is 17.2 years and standard deviation is 3.45 years.

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Q: Problem 4 In the last 57 matches Akram's batting average and standard deviation of runs

A: Given, n1 = 57        x¯1 = 25       σ1 = 3n2 = 39         x¯2 = 20       σ2 = 5         To find, a.…

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A: Given that.   X~N( μ , ?)   μ=280     , ?=60   Z-score =( x - μ )/?

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Q: the mean is 146, and standard deviation is 35. A score of 181 is how many z-scores above the mean?

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A: The mean is 180 acres.

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A: GivenMean(μ)=11.6standard deviation(2.8)

Q: Question 20 only, please

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Q: In 1940 the average size of a U.S. farm was 174 acres. Let's say that the standard deviation was 58…

A: Given information Mean (µ) = 174 acres Standard deviation (σ) = 58 acres n =41 Middle 50% is given…

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Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.