This is a standard Normal Distribution application Problem. I want you to show me how they got the final answer by solving equation 1 and 2. See the attached solution.QuestionThe weights of boxes of oranges are normally distributed such that 30% of them are greater than 4kgand 20% are greater than 4.53kg. Estimate the mean and standard deviation of the weights. We are given that P(X >4) = 0.3 and P(X >4.53) = 0.2. Using this we have to find the values of u and0. Lets first consider P(X >4) = 0.3. Converting 4 into z value we get 1-4, so we have4-uP(X >4) = P| z>.= 0.3. Now draw a standard normal curve and shade the area on the rightofto be 0.3.0.34- 4Using the tables we obtain the z value to be 0.52. Therefore, we have the equation4-H0.52 =(1)Similarly using P(X >4.53) = 0.2 so we get another equation4.53 - 40.84 =(2)Solving the equations (1) and (2) simultaneously we get p = 3.12 kg and o = 1.68 kg.

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Answered: This is a standard Normal Distribution…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

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This is a standard Normal Distribution application Problem.

I want you to show me how they got the final answer by solving equation 1 and 2.

See the attached solution.

Question

The weights of boxes of oranges are normally distributed such that 30% of them are greater than 4kg
and 20% are greater than 4.53kg. Estimate the mean and standard deviation of the weights.

We are given that P(X >4) = 0.3 and P(X >4.53) = 0.2. Using this we have to find the values of u and
0. Lets first consider P(X >4) = 0.3. Converting 4 into z value we get 1-4, so we have
4-u
P(X >4) = P| z>.
= 0.3. Now draw a standard normal curve and shade the area on the right
of
to be 0.3.
0.3
4- 4
Using the tables we obtain the z value to be 0.52. Therefore, we have the equation
4-H
0.52 =
(1)
Similarly using P(X >4.53) = 0.2 so we get another equation
4.53 - 4
0.84 =
(2)
Solving the equations (1) and (2) simultaneously we get p = 3.12 kg and o = 1.68 kg.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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