Answer:
Step-by-step explanation:
Hello!
X: number of absences per tutorial per student over the past 5 years(percentage)
X≈N(μ;σ²)
You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.
The formula for the CI is:
X[bar] ± 
* 
⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.
Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4
X[bar]= 10.41
S= 3.71
[10.41±1.645*
]
[9.26; 11.56]
Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.
I hope this helps!
First we have to figure out how many hours have passed. Since 1PM also means 1300 hours, we'll subtract 9 from 13. 13-9=4. therefore, 4 hours have passed. if it gains 4 minutes every hour, we'll multiply the amount of hours that have passed by the amount the clock gains every hour. 4x4=16. now we subtract 16 from 1300 because the clock is ahead of the real time, but remember and hour only has 60 minutes! 1300-16=1244. Therefore, the correct time is 12:44PM.
Answer:
Step-by-step explanation:
Given
Required
Solve for s
Multiply both sides by 4
Open bracket
Subtract 300 from both sides
Divide both sides by 3
Split
Answer:
0.002
Step-by-step explanation:
We need to estimate the standard error of the mean, so we can use it as a standard deviation of the sample of 50 males.
Standard error of the mean = standard deviation/√n
Standard error of the mean = 32/√50 = 4.52
Now we can use this Standard error of the mean to estimate z as follows:
Z = (x – mean)/standard deviation
Z = (190-177)/4.52
Z = 2.87
Using a Z table we can find probability that mean is under 190
P (z<190)= 0.998
For the probability that the mean exceed 190 lbs we substract from 1
P(z>190) = 1 - 0.998 = 0.002
