Answer: 0.75
Step-by-step explanation:
Given : Interval for uniform distribution : [0 minute, 5 minutes]
The probability density function will be :-
The probability that a given class period runs between 50.75 and 51.25 minutes is given by :-
![P(x>1.25)=\int^{5}_{1.25}f(x)\ dx\\\\=(0.2)[x]^{5}_{1.25}\\\\=(0.2)(5-1.25)=0.75](https://tex.z-dn.net/?f=P%28x%3E1.25%29%3D%5Cint%5E%7B5%7D_%7B1.25%7Df%28x%29%5C%20dx%5C%5C%5C%5C%3D%280.2%29%5Bx%5D%5E%7B5%7D_%7B1.25%7D%5C%5C%5C%5C%3D%280.2%29%285-1.25%29%3D0.75)
Hence, the probability that a randomly selected passenger has a waiting time greater than 1.25 minutes = 0.75
Complete question :
Wright et al. [A-2] used the 1999-2000 National Health and Nutrition Examination Survey NHANES) to estimate dietary intake of 10 key nutrients. One of those nutrients was calcium in all adults 60 years or older a mean daily calcium intake of 721 mg with a standard deviation of 454. Usin these values for the mean and standard deviation for the U.S. population, find the probability that a randonm sample of size 50 will have a mean: (mg). They found a) Greater than 800 mg b) Less than 700 mg. c) Between 700 and 850 mg.
Answer:
0.10935
0.3718
0.9778
0.606
Step-by-step explanation:
μ = 721 ; σ = 454 ; n = 50
P(x > 800)
Zscore = (x - μ) / σ/sqrt(n)
P(x > 800) = (800 - 721) ÷ 454/sqrt(50)
P(x > 800) = 79 / 64.205295
P(x > 800) = 1.23
P(Z > 1.23) = 0.10935
2.)
Less than 700
P(x < 700) = (700 - 721) ÷ 454/sqrt(50)
P(x < 700) = - 21/ 64.205295
P(x < 700) = - 0.327
P(Z < - 0.327) = 0.3718
Between 700 and 850
P(x < 850) = (850 - 721) ÷ 454/sqrt(50)
P(x < 850) = 129/ 64.205295
P(x < 700) = 2.01
P(Z < 2.01) = 0.9778
P(x < 850) - P(x < 700) =
P(Z < 2.01) - P(Z < - 0.327)
0.9778 - 0.3718
= 0.606
Answer:
x>0
Step-by-step explanation:
Answer:
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