Exact form is 469/12 decimal form is 39.083 repeating and mixed number form is 39/1/12
Answer:
Kindly check explanation
Step-by-step explanation:
H0 : μ = 5500
H1 : μ > 5500
The test statistic assume normal distribution :
Test statistic :
(Xbar - μ) ÷ s/sqrt(n)
(5625.1 - 5500) ÷ 226.1/sqrt(15) = 2.1429 = 2.143
Pvalue from test statistic :
The Pvalue obtained using the calculator at df = 15 - 1 = 14 is 0.025083
α = 0.05
Since ;
Pvalue < α
0.025083 < 0.05 ; Reject H0
The confidence interval :
Xbar ± Tcritical * s/sqrt(n)
Tcritical at 95% = 1.761 ;
margin of error = 1.761 * 226.1/sqrt(15) = 102.805
Lower boundary : (5625.1 - 102.805) = 5522.295
(5522.295 ; ∞)
The hypothesized mean does not occur within the constructed confidence boundary. HENCE, there is significant eveidbce to support the claim that the true mean life of a biomedical device is greater than 5500
The standard form of a line:
![Ax + By = C](https://tex.z-dn.net/?f=Ax%20%2B%20By%20%3D%20C)
![\dfrac{1}{8}y=-\dfrac{3}{4}x-1\ \ \ \ |\text{multiply both sides by 8}\\\\\not8^1\cdot\dfrac{1}{\not8_1}y=\not8^2\cdot\left(-\dfrac{3}{\not4_1}\right)x-1\\\\y=2(-3x)-1\\\\y=-6x-1\ \ \ \ |\text{add 6x to both sides}\\\\\boxed{6x+y=-1}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B8%7Dy%3D-%5Cdfrac%7B3%7D%7B4%7Dx-1%5C%20%5C%20%5C%20%5C%20%7C%5Ctext%7Bmultiply%20both%20sides%20by%208%7D%5C%5C%5C%5C%5Cnot8%5E1%5Ccdot%5Cdfrac%7B1%7D%7B%5Cnot8_1%7Dy%3D%5Cnot8%5E2%5Ccdot%5Cleft%28-%5Cdfrac%7B3%7D%7B%5Cnot4_1%7D%5Cright%29x-1%5C%5C%5C%5Cy%3D2%28-3x%29-1%5C%5C%5C%5Cy%3D-6x-1%5C%20%5C%20%5C%20%5C%20%7C%5Ctext%7Badd%206x%20to%20both%20sides%7D%5C%5C%5C%5C%5Cboxed%7B6x%2By%3D-1%7D)
Step-by-step explanation:
n-4≥13
n≥13+4
n≥17
point the no. 17 in the number line.
Answer:
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Let X the random variable who represent the exam scores for the population, and for this case we know the distribution for X is given by:
Where
and
And let
represent the sample mean, the distribution for the sample mean is given by:
What is the probability that a randomly selected score will be between 69 and 77?
For this case we can use the z score formula given by:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
And that correspond with the 35.75% of the data.