The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard
1 answer:
Answer:
0.6563 or 65.63% of brook trout caught will be between 12 and 18 inches
Step-by-step explanation:
Mean trout length (μ) = 14 inches
Standard deviation (σ) = 3 inches
The z-score for any given trout length 'X' is defined as:
e interval
For a length of X =12 inches:
According to a z-score table, a score of -0.6667 is equivalent to the 25.25th percentile of the distribution.
For a length of X =18 inches:
According to a z-score table, a score of 1.333 is equivalent to the 90.88th percentile of the distribution.
The proportion of trout caught between 12 and 18 inches, assuming a normal distribution, is the interval between the equivalent percentile of each length:
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<h2>Answer:</h2><h3>(0,14) and (2,11)</h3><h2>Step-by-step explanation:</h2>
points are on the graph of the line:-
⠀
we get,
⠀
⠀
In,(0,14)
If we put the value ,x=0
y=
So,we get one point,(0,14)✔
In,(2,11)
If we put ,x=2
y=
so,we get another point (2,11)✔
in other given ponts{(1,12),(4,9),(8,4)},It is impossible. ⠀ ⠀
<h2>Answer with explanation:</h2>
Given : In a restaurant, the proportion of people who order coffee with their dinner is p.
Sample size : n= 144
x= 120
The null and the alternative hypotheses if you want to test if p is greater than or equal to 0.85 will be :-
Null hypothesis : [ it takes equality (=, ≤, ≥) ]
Alternative hypothesis : [its exactly opposite of null hypothesis]
∵Alternative hypothesis is left tailed, so the test is a left tailed test.
Test statistic :
Using z-vale table ,
Critical value for 0.05 significance ( left-tailed test)=-1.645
Since the calculated value of test statistic is greater than the critical value , so we failed to reject the null hypothesis.
Conclusion : We have enough evidence to support the claim that p is greater than or equal to 0.85.