Answer:
the number of families needed for the survey is 2,157
Step-by-step explanation:
The computation of the number of families needed for the survey is shown below:
Given that
The Standard deviation is $900
Service level = 99%
z value at 99% = 2.58
Expenditures = $50
Based on the above information
The number of families needed for the survey is
= 2,157
Hence, the number of families needed for the survey is 2,157
Yes. The 100 - 85 = 15, meaning 15% of parents said yes. 85 > 15
Answer:
36
Step-by-step explanation:
180-144=36
Answer:
144
Step-by-step explanation:
so if we have 120 and we need to find out what the original prices we could add 20% of 120 which will get us to 144
Answer:
a) x = 119 minutes
the mean value for average movie length in minutes is 119 minutes
b) margin of error M.E = 44 minutes
Note; Since the number of samples used is not given, the standard deviation r cannot be calculated using the equation
M.E = zr/√n
Step-by-step explanation:
Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.
The confidence interval of a statistical data can be written as.
x+/-zr/√n
x+/-M.E
M.E = zr/√n
Given that;
M.E = margin of error
Mean = x
Standard deviation = r
Number of samples = n
Confidence interval = 95%
z value (at 95% confidence) = 1.96
a) The mean value x can be calculated as;
x = (a+b)/2
Where a and b are the lower and upper bounds of the confidence interval;
a = 75 minutes
b = 163 minutes
substituting the values;
x = (75+163)/2
x = 119 minutes
the mean value for average movie length in minutes is 119 minutes
b) the margin of error M.E can be calculated as;
M.E = (b-a)/2
Substituting the values;
M.E = (163-75)/2
M.E = 44 minutes
Since the number of samples used is not given, the standard deviation r cannot be calculated using the equation
M.E = zr/√n
