Answer:
And rounded up we have that n=656
Step-by-step explanation:
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 80% of confidence, our significance level would be given by
and
. And the critical value would be given by:
Solution to the problem
The margin of error for the proportion interval is given by this formula:
(a)
Since we don't have prior info for the proportion of interest we can use
as estimator. And on this case we have that
and we are interested in order to find the value of n, if we solve n from equation (a) we got:
(b)
And replacing into equation (b) the values from part a we got:
And rounded up we have that n=656
Answer:boys: median=50, mean=46
Girls: median= 33, mean=34
Step-by-step explanation:
Mean is the average. Median is the middle number. Both groups had 10 scores. Boys ranged from 15-81. Girls from 0-75; their mean & median would be less because their scores are less.
(8 - 3i)(2 - 7i)=
16 x -56i - 6i + 21i^2
16 x -50i - 21
Answer:
cheat off somebody who has the answer
Step-by-step explanation:
Answer:
the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds is 0.0166
Step-by-step explanation:
The summary of the given statistical data set are:
Sample Mean = 186
Standard deviation = 29
Maximum capacity 3,417 pounds or 17 persons.
sample size = 17
population mean =3417
The objective is to determine the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds
In order to do that;
Let assume X to be the random variable that follows the normal distribution;
where;
Mean
= 186 × 17 = 3162
Standard deviation = 
Standard deviation = 119.57






Therefore; the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds is 0.0166