A distribution of values is normal with a mean of 188.4 and a standard deviation of 57. Find the probability that a randomly selected value is greater than 120. P(X 120) = Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted. Question Help: Video

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A: given data, normal distribution , μ=100σ=15we have to find out the shaded region for the given…

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A: From the provided information, Mean (µ) = 75 Standard deviation (σ) = 20 X~N (75, 20)

Q: The area of the shaded region is

A: Consider X denotes the IQ score. X~Nμ=100,σ=15

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A: Given that  The data is normally distributed with  Mean (μ) = 100 Standard deviation (σ) = 15   We…

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Q: Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those…

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Q: Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those…

A: normal distribution μ = 100σ = 15 P(85<x<110) = ?

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A: For January, mean = 36 and SD = 9 Fir July, mean = 77 and SD = 8 First we calucuate the z scores of…

Q: Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those…

A: The graph depicts IQ scores of adultsThe scores are normally distributed.Mean () = 100Standard…

Q: Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those…

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A: Given that,The middle 68% of the scores fall between 77 and 93.

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A: Given that  Population mean μ =5 feet 6.3 inches =66.3 1feet=12 inches  5feet=12×5=60+6.3=66.3…

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A: Sample mean, x¯ = 98.382 sample standard deviation = s = 0.712 sample size, n = 50 Here the sample…

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A: From the given information we find the range and standard deviation.

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