The lower and upper limit of the 95% confidence interval is 183.2 and 196.0 unit.
What is a confidence interval ?
When you run your experiment again or resample the population in the same way, you can expect your estimate to fall within a certain range of values a certain percentage of the time. This is known as the confidence interval.
Main Body:
Given in question :
x- bar = 189.6
standard deviation = 10
C.I. = 95% =0.95
N=12
so, degree of freedom (D.F) = N-1 = 11
confidence level (α) = (1- C.I.)/2
= 0.025
according to t- distribution table for d.f = 11 and α= 0.025 , my result is 2.201
Now dividing standard deviation by square root of samples, we get
= 10/√12= 2.887
Lower limit = Mean - 2.201*2.887
= 189.6- 6.354
= 183.246
Upper limit = Mean + 2.201*2.887
= 189.6 + 6.354
= 195.954
Hence the result is 183.2 and 196.
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Answer:
38.4
Step-by-step explanation:
(multiply) hkssfejshwjwjgwskfhkdshejkdhdc (don't mind the random letters it needed to be 20 characters)
The measure of ∠DBC is 36°
In the circle, BD is the diameter and BC is the minor chord on the circle, due to which ∠BCD = 90°, in triangle BDC
Then,
Given arc BD measure is 54° i.e ∠BDC is 54°
The, from property of triangle, sum of all angles is 180°, we can find the third angle i.e. ∠DBC.
Thus, ∠DBC = 180°- 90°- 54°
∠DBC = 36°
Answer:
We know that n = 50 and p =0.78.
We need to check the conditions in order to use the normal approximation.
Since both conditions are satisfied we can use the normal approximation and the distribution for the proportion is given by:
![p \sim N (p, \sqrt{\frac{p(1-p)}{n}})](https://tex.z-dn.net/?f=%20p%20%5Csim%20N%20%28p%2C%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%29%20)
With the following parameters:
![\mu_ p = 0.78](https://tex.z-dn.net/?f=%20%5Cmu_%20p%20%3D%200.78)
![\sigma_p = \sqrt{\frac{0.78*(1-0.78)}{50}}= 0.0586](https://tex.z-dn.net/?f=%5Csigma_p%20%3D%20%5Csqrt%7B%5Cfrac%7B0.78%2A%281-0.78%29%7D%7B50%7D%7D%3D%200.0586)
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
We know that n = 50 and p =0.78.
We need to check the conditions in order to use the normal approximation.
Since both conditions are satisfied we can use the normal approximation and the distribution for the proportion is given by:
![p \sim N (p, \sqrt{\frac{p(1-p)}{n}})](https://tex.z-dn.net/?f=%20p%20%5Csim%20N%20%28p%2C%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%29%20)
With the following parameters:
![\mu_ p = 0.78](https://tex.z-dn.net/?f=%20%5Cmu_%20p%20%3D%200.78)
![\sigma_p = \sqrt{\frac{0.78*(1-0.78)}{50}}= 0.0586](https://tex.z-dn.net/?f=%5Csigma_p%20%3D%20%5Csqrt%7B%5Cfrac%7B0.78%2A%281-0.78%29%7D%7B50%7D%7D%3D%200.0586)
Answer:
25
Step-by-step explanation:
Convert fraction (ratio) 6 / 24 Answer: 25%