Answer:
75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5 pounds.
Step-by-step explanation:
The question is missing. It is as follows:
Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains had the following weights (pounds): 69 104 125 129 60 64
Assume that the population of x values has an approximately normal distribution.
Find a 75% confidence interval for the population average weight μ of all adult mountain lions in the specified region. (Round your answers to one decimal place.)
75% Confidence Interval can be calculated using M±ME where
- M is the sample mean weight of the wild mountain lions (
)
- ME is the margin of error of the mean
And margin of error (ME) of the mean can be calculated using the formula
ME=
where
- t is the corresponding statistic in the 75% confidence level and 5 degrees of freedom (1.30)
- s is the standard deviation of the sample(31.4)
Thus, ME=
≈16.66
Then 75% confidence interval is 91.8±16.66. That is between 75.1 and 108.5
Hello there. :)
<span>How many miles are in 400 kilometers
248.548
</span>
Answer:
Step-by-step explanation:
P(a or b) = P(a) + P(b) - P(a and b)
0.88 = 0.70 + 0.60 - P(a and b)
P(a and b) = 1.30 - 0.88 = 0.42.
So the events a and b are not mutually exclusive.
Answer:
7 units
Step-by-step explanation:
<span>To determine whether (−1,4) is a solution to the equation 3x+8y=29, substitute ...-1... for x and ...4... and for y.
In a pair of values like (-1, 4), the first coordinate, or entry, occupied by -1 always represents x.
Similarly, the second coordinate occupied by 4 represents y.
Answer: </span>substitute ...-1... for x and ...4... and for y.