Continuous random variable, X, has anuIKnOwn probablity density function.The table below gives the area underneath theprobablity density function to the right of specificvalues of x.0.00.10.20.30.40.50.60.70.80.91.00.84648 0.83249 0.81873 0.80520 0.79189 0.778800.76593 0.75327 0.74082 0.728572.00.71653 0.70469 0.69304 0.68159 0.67032 0.65924 0.64834 0.63763 0.62709 0.616723.00.60653 0.59651 0.586650.57695 0.56741 0.55804 0.54881 0.53974 0.53082 0.522054.00.51342 0.50493 0.49659 0.488380.48031 0.47237 0.46456 0.45688 0.44933 0.441905.00.43460 0.42742 0.420350.41340 0.40657 0.399850.39324 0.38674 0.38035 0.37406Use the table above to find:(A)P(x < 2.1) =(B)P(x > 1.57) =

Since, the table given gives the area underneath the probability density function to the right of…

Answered: Continuous random variable, X, has an…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A Moving to the next question prevents changes to this answer. tion 9 Elongation (in %) of steel plates treated with aluminum is a random variable with the probability density function for 18 < x < 24 f(x) = {126 otherwise Find the variance of the elongation. O 2.980 O 8.222 5.297 O 3.190 A Moving to the next question prevents changes to this answer. MacBoo
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Draw the standard normal curve that corresponds to an area between z = 0 and z = 2. Then, use the standard normal table to find the area under the curve between these two z-scores.
Below is a graph of a normal distribution with mean =μ1 and standard deviation =σ4. The shaded region represents the probability of obtaining a value from this distribution that is between 3and 5. 0.1 0.2 0.3 0.4 X 1 3 5 Shade the corresponding region under the standard normal density curve below. 0.1 0.2 0.3 0.4 Z 1 2 3 4 5 6 7 8 9 -1 -2 -3 -4 -5 -6 -7 -8 -9
Solve for D
Let x be a continuous random variable that follows a normal distribution with a mean of 200 and a standard deviation of 25. Find the value of x so that the area under the normal curve to the right of x is approximately 0.05.
Find the area under the standard normal curve to the right of z = 2.04 z= 1.09 Genetics: Pea plants contain two genes for seed color, each of which may be Y (for yellow seeds) or G (for green seeds). Plants that contain one of each type of gene are called heterozygous. According to the Mendelian theory of genetics, if two heterozygous plants are crossed, each of their offspring will have probability 0.75 of having yellow seeds and probability 0.25 of having green seeds. One hundred such offspring are produced. Approximate the probability that more than 30 have green seeds.
Determine the area under the standard normal curve that lies to the right of the z-score 0.05 and to the left of the z-score 0.25. z−0.2−0.10.00.10.20.30.40.50.000.42070.46020.50000.53980.57930.61790.65540.69150.010.41680.45620.50400.54380.58320.62170.65910.69500.020.41290.45220.50800.54780.58710.62550.66280.69850.030.40900.44830.51200.55170.59100.62930.66640.70190.040.40520.44430.51600.55570.59480.63310.67000.70540.050.40130.44040.51990.55960.59870.63680.67360.70880.060.39740.43640.52390.56360.60260.64060.67720.71230.070.39360.43250.52790.56750.60640.64430.68080.71570.080.38970.42860.53190.57140.61030.64800.68440.71900.090.38590.42470.53590.57530.61410.65170.68790.7224 Use the value(s) from the table above.

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