Suppose that the number of insects captured in a trap on different nights is normally distributed with mean 2950 and standard deviation 550.
b) Suppose that we change the location of a trap whenever it captures a
number of insects that fall in the lowest 5%. If a trap was relocated on a given night, what is the maximum number of insects that it captured?
**I don't understand what this question is asking... is there another way to word this (in terms of what i'm trying to find here?)**
They are just asking you for the "cutoff" number that is required to avoid having to move the trap. It will be the number such that 95% of the time, a larger number is caught. You get that number by integrating the normal distribution, from zero insects to whatever number gives you an integral of 5% of the total distribution. Using the tool at this website,
http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html ,
I get the answer to be 2045
Good Afternoon,
I need to subtract the raction 7 4/5 from the fraction 18 6/15 and it needs to be reduced the the lowest terms! help
Good Afternoon,
I need to subtract the fraction 7 4/5 from the fraction 18 6/15 and it needs to be reduced the the lowest terms! help
To subtract fractions, you first need to find a common denominator. In this case, the common denominator is 15.
Now let's convert the mixed numbers into improper fractions:
18 6/15 = (18 * 15 + 6) / 15 = 276/15
7 4/5 = (7 * 5 + 4) / 5 = 39/5
Now, we can subtract the fractions:
276/15 - 39/5
To subtract fractions, we need a common denominator, which is 15 in this case.
276/15 - 39/5 = (276 * 1) / (15 * 1) - (39 * 3) / (5 * 3)
= 276/15 - 117/15
Now, we can subtract the numerators and keep the common denominator:
(276 - 117) / 15 = 159/15
To simplify the fraction to lowest terms, we can divide both the numerator and denominator by their greatest common divisor. In this case, the greatest common divisor of 159 and 15 is 3.
159/15 = (159 ÷ 3) / (15 ÷ 3) = 53/5
So the answer is 53/5 in lowest terms.