Answer:
The 95% confidence interval for the average credit score is between 697.675 and 802.325
Step-by-step explanation:
We are in posession of the sample standard deviation, so the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 20 - 1 = 19
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.093
The margin of error is:
M = T*s = 2.093*25 = 52.325
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 750 - 52.325 = 697.675
The upper end of the interval is the sample mean added to M. So it is 750 - 52.325 = 802.325
The 95% confidence interval for the average credit score is between 697.675 and 802.325
Answer:
= 100
Step-by-step explanation:
The number of visitors is modeled by the expression
Where m is the number of months after the number of visitors was measured.
We need to evaluate the above expression for m = -2.
So,
At m = -2, the value of the above expression is equal to 100.
Answer:
Move downward by 7 units
Move leftward by 6 units
Step-by-step explanation:
Given
See attachment for grid
Required
The transformation from the current location to the new location
To do this, we pick two corresponding points on the current location and the new location.
We have:
-- Current location
-- New location
First, move A downwards by 7 units.
The rule to this is:
So, we have:
Next, move the above points leftward by 6 units.
The rule to this is:
So, we have:
Answer:
Step-by-step explanation:
hello : look this solution
Answer:
The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors
P(X⁻ ≥ 96.3) = 0.0087
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that the mean of the Population = 95
Given that the standard deviation of the Population = 5
Let 'X' be the random variable in a normal distribution
Let X⁻ = 96.3
Given that the size 'n' = 84 monitors
<u><em>Step(ii):-</em></u>
<u><em>The Empirical rule</em></u>
Z = 2.383
The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors
P(X⁻ ≥ 96.3) = P(Z≥2.383)
= 1- P( Z<2.383)
= 1-( 0.5 -+A(2.38))
= 0.5 - A(2.38)
= 0.5 -0.4913
= 0.0087
<u><em>Final answer:-</em></u>
The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors
P(X⁻ ≥ 96.3) = 0.0087