Missing question: How many executives should be surveyed?
Solution:
Mean = 13 hours
SD = 3 hours
Confidence level = 95%
Mean viewing time within a quarter of an hour = 0.25 = 1.96*3/sqrt (N)
Where N = Sample population
N = {(1.96*3)/0.25}^2 = 553.19 ≈ 554
Therefore, 554 executives should be surveyed to yield such results.
Answer:
The limit that 97.5% of the data points will be above is $912.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

Find the limit that 97.5% of the data points will be above.
This is the value of X when Z has a pvalue of 1-0.975 = 0.025. So it is X when Z = -1.96.
So




The limit that 97.5% of the data points will be above is $912.
Answer:x = 9/16
Step-by-step explanation:
2x + 3 = 6x - 4
/3
First you gotta get rid of the divided by 3
2x / 3 = 2/3x and 3 / 3 = 1
Now you have
2/3x + 1 = 6x - 4
Second you have to get rid of the 2/3x on the left side
2/3x + 1 = 6x - 4
-2/3x -2/3x
1 = 5 1/3x - 4
Third you have to get rid of the -4
1 = 5 1/3x - 4
-4 -4
-3 = 5 1/3x
Fourth you have to divide by 5 1/3
-3 = 5 1/3x
/ 5 1/3 / 5 1/3
x = 9/16