Answer:
The standard deviation of the sample mean differences is _5.23_
Step-by-step explanation:
We have a sample of a population A and a sample of a population B.
For the sample of population A, the standard deviation is
The sample size is:
.
For the sample of population B, the standard deviation is
The sample size is:
.
Then the standard deviation for the difference of means has the following form:
Finally
Answer:
88.88% probability that it endures for less than a year and a half
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 question, we have that:
The next career begins on Monday; what is the likelihood that it endures for less than a year and a half?
One year has 52.14 weeks. So a year and a half has 1.5*52.14 = 78.21 weeks.
So this probability is the pvalue of Z when X = 78.21.
has a pvalue of 0.8888
88.88% probability that it endures for less than a year and a half
Answer:
Step-by-step explanation:
Let x = the number of all the species of birds sighted in the park
“300 species of birds that migrate. This accounts for 2/7 of all the species of birds sighted in the park”
the above sentence means “(2/7)x = 300”
now we can solve it :
(2/7)x = 300
⇔ 2x = 7×300
⇔ 2x = 2100
⇔ x = 2100/2
⇔ x = 1050