Assuming that it is known from previous studies that σ = 4.5 grams, how many mice should be included in our sample if we wish to
1 answer:
Answer:
The required amount of mice is 11.82 gram.
Step-by-step explanation:
Given : Assuming that it is known from previous studies that σ = 4.5 grams. If we wish to be 95% confident that the mean weight of the sample will be within 3 grams of the population mean for all mice subjected to this protein diet.
To find : How many mice should be included in our sample?
Solution :
The formula used in the situation is
Where, z value at 95% confidence interval is z=1.96
gram is the mean of the population
gram is the standard deviation of the sample
Substituting the value, to find x sample mean
Therefore, The required amount of mice is 11.82 gram.
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Answer:
Step-by-step explanation:
Total number of people = 130
Number of people who use the gym = 73
Number of people who use the pool = 62
Number of people who use the track = 58
Number of people who use the gym and the pool = 22
Number of people who use the pool and the track = 29
Number of people who use the gym and the track = 25
Number of people who use all three facilities = 11
Total number of people who use at least two facilities = 22 + 29 + 25 + 11 = 87
The probability that the randomly selected person uses all three facilities = number of those who use all three facilities ÷ total number of people who use at least two facilities.
==> 11 ÷ 87
==>
32 × 2 = 68
32 × 3 = 96
32 × 4 = 128
So far we can conclude that the answer to your question lies somewhere between the numbers 3 and 4. To narrow down the answer some more, multiply 32 by 3.5 (a midway point between 3 and 4).
32 × 3.5 = 112
The number 112 tells us that the decimal we are looking for is higher than 3.5. (Because we need to get to 125, not 112.) Let's try some decimals between 3.5 and 4.
32 × 3.7 = 118.4
32 × 3.8 = 121.6
32 × 3.9 = 124.8
32 × 4 = 128
As we narrow down our answer, we can see that the number we are looking for lies between 3.9 and 4 on the number line. Now we need to start testing some decimals between 3.9 and 4.
32 × 3.905 = 124.96
Again, use the number five as a "midway" point to decide if you should use numbers that are higher or lower than 3.905. In this case, we need to use numbers higher than 3.905.
32 × 3.906 = 124.992
32 × 3.907 = 125.024
We are getting even closer to our number now that we know the decimal is somewhere between 3.906 and 3.907.
32 × 3.9065 = 125.008
With our midway point we can see that our number lies between 3.906 and 3.9065. Let's try a quarter point to see where our number lies from there.
32 × 3.90625 = 125
And BINGO! We have found the answer to the question. To be rephrased, our answer can be put like this:
= 3.90625
32 × 3 = 96
32 × 4 = 128
So far we can conclude that the answer to your question lies somewhere between the numbers 3 and 4. To narrow down the answer some more, multiply 32 by 3.5 (a midway point between 3 and 4).
32 × 3.5 = 112
The number 112 tells us that the decimal we are looking for is higher than 3.5. (Because we need to get to 125, not 112.) Let's try some decimals between 3.5 and 4.
32 × 3.7 = 118.4
32 × 3.8 = 121.6
32 × 3.9 = 124.8
32 × 4 = 128
As we narrow down our answer, we can see that the number we are looking for lies between 3.9 and 4 on the number line. Now we need to start testing some decimals between 3.9 and 4.
32 × 3.905 = 124.96
Again, use the number five as a "midway" point to decide if you should use numbers that are higher or lower than 3.905. In this case, we need to use numbers higher than 3.905.
32 × 3.906 = 124.992
32 × 3.907 = 125.024
We are getting even closer to our number now that we know the decimal is somewhere between 3.906 and 3.907.
32 × 3.9065 = 125.008
With our midway point we can see that our number lies between 3.906 and 3.9065. Let's try a quarter point to see where our number lies from there.
32 × 3.90625 = 125
And BINGO! We have found the answer to the question. To be rephrased, our answer can be put like this: