Answer:
0.3333 = 33.33% probability that the employee will arrive between 8:15 a.m. and 8:25 a.m.
Step-by-step explanation:
A distribution is called uniform if each outcome has the same probability of happening.
The uniform distributon has two bounds, a and b, and the probability of finding a value between c and d is given by:
A particular employee arrives at work sometime between 8:00 a.m. and 8:30 a.m.
We can consider 8 am = 0, and 8:30 am = 30, so 
Find the probability that the employee will arrive between 8:15 a.m. and 8:25 a.m.
Between 15 and 25, so:
0.3333 = 33.33% probability that the employee will arrive between 8:15 a.m. and 8:25 a.m.
Answer:
non linear ( ꈍᴗꈍ)( ╹▽╹ )( ╹▽╹ )
Answer:
standard form for a quadratic expression
Step-by-step explanation:
This is a quadratic expression. If you want this expression in standard form, write the terms in descending order of powers of x: -2x^2 + 1.
This -2x^2 + 1 has not yet been factored.
-2x^2 + 1 is in standard form.
The mean of the given sample data is 210, and the standard deviation is 7.937.
Given size 'n' = 300
The population proportion 'p' = 0.7
Let 'x' be the random variable of the binomial distribution
a) mean of the binomial distribution = n p = 300 × 0.7
μ = 210
b) variance of the binomial distribution
⇒ n p q
⇒ 300 × 0.7 ×0.3
⇒ σ² = 63
The standard deviation of the binomial distribution:
⇒ √n p q = √63 = 7.937
Thus, the mean of the given sample data is 210, and the standard deviation is 7.937.
Learn more about the standard deviation here:
brainly.com/question/16555520
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The question seems to be incomplete the correct question would be:
describe the sampling of p hat. Assume that the size of the population is 25000 n= 300 p=0.7 a) Determine the mean of the sampling distributionb) Dtermine the standard deviation of the sampling distribution
