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13.02.2020 •
Mathematics
A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100 weights is taken.
a. What is the distribution for the weights of one 25-pound lifting weight? What is the mean and standard deviation?
b. What is the distribution for the mean weight of 100 25-pound lifting weights?
c. Find the probability that the mean actual weight for the 100 weights is less than 24.9.
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Ответ:
a) Mean = 25 , S.d = 0.5774
b) X~N(25 , 0.05774)
c) P (X < 24.9) = 0.0416
Step-by-step explanation:
Given:
- Limits of a uniform distribution are:
U( 24 , 26 )
- Sample size n= 100
Find:
a. What is the distribution for the weights of one 25-pound lifting weight? What is the mean and standard deviation?
Solution
The mean and standard deviation value is:
E(X) = ( 24 + 26 ) / 2 = 25
Var(X) = (25 - 24)/3 = 1/3
s.d(X) = sqrt(1/3) = 0.5774
Find:
b. What is the distribution for the mean weight of 100 25-pound lifting weights?
Solution
The random variable X follows a normal distribution:
X~N(25 , s.d/100)
X~N(25 , 0.05774)
Find:
c. Find the probability that the mean actual weight for the 100 weights is less than 24.9.
Solution
The probability of P(X < 24.9) is:
P (X < 24.9) = P ( Z < (24.9-25)/0.05774)
= P(Z < -1.7319)
= 0.0416
Ответ:
Well, 6 x 90 = 540
So to answer your question, there is only one zero.