Respuesta :
Answer:
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
Using the uniform distribution, the normal distribution and the Central Limit Theorem, it is found that:
a) The distribution is uniform, with mean of 25 pounds and standard deviation of 0.5774 pounds.
b) The distribution is approximately normal, with mean of 25 pounds and standard error of 0.05774 pounds.
c) There is a 0.0418 = 4.18% probability that the mean actual weight for the 100 weights is less than 24.9.
A uniform distribution has two bounds, a and b.
It's mean is given by:
[tex]E(X) = \frac{a + b}{2}[/tex]
It's standard deviation is given by:
[tex]S(X) = \sqrt{\frac{(b - a)^2}{12}}[/tex]
In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
Item a:
Uniform between 24 and 26 pounds, hence [tex]a = 24, b = 26[/tex].
The mean is:
[tex]E(X) = \frac{a + b}{2} = \frac{24 + 26}{2} = 25[/tex]
The standard deviation is:
[tex]S(X) = \sqrt{\frac{(b - a)^2}{12}} = \sqrt{\frac{(26 - 24)^2}{12}} = 0.5774[/tex]
The distribution is uniform, with mean of 25 pounds and standard deviation of 0.5774 pounds.
Item b:
Sample of 100, hence [tex]n = 100, s = \frac{0.5774}{\sqrt{100}} = 0.05774[/tex].
The distribution is approximately normal, with mean of 25 pounds and standard error of 0.05774 pounds.
Item c:
This probability is the p-value of Z when X = 24.9, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{24.9 - 25}{0.05774}[/tex]
[tex]Z = -1.73[/tex]
[tex]Z = -1.73[/tex] has a p-value of 0.0418.
There is a 0.0418 = 4.18% probability that the mean actual weight for the 100 weights is less than 24.9.
A similar problem is given at https://brainly.com/question/24663213
