A study indicates that 37% of students have laptops. You randomly sample 30 students. Find the mean and the standard deviation o
1 answer:
Answer:
The mean and the standard deviation of the number of students with laptops are 1.11 and 0.836 respectively.
Step-by-step explanation:
Let <em>X</em> = number of students who have laptops.
The probability of a student having a laptop is, P (X) = <em>p</em> = 0.37.
A random sample of <em>n</em> = 30 students is selected.
The event of a student having a laptop is independent of the other students.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.
The mean and standard deviation of a binomial random variable <em>X</em> are:
Compute the mean of the random variable <em>X</em> as follows:
The mean of the random variable <em>X</em> is 1.11.
Compute the standard deviation of the random variable <em>X</em> as follows:
The standard deviation of the random variable <em>X</em> is 0.836.
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Answer:
About 0.6548 grams will be remaining.
Step-by-step explanation:
We can write an exponential function to model the situation. The standard exponential function is:
The original sample contained 510 grams. So, a = 510.
Each half-life, the amount decreases by half. So, r = 1/2.
For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.
Therefore, our function is:
One year has 365 days.
Therefore, the amount remaining after one year will be:
About 0.6548 grams will be remaining.
Alternatively, we can use the standard exponential growth/decay function modeled by:
The starting sample is 510. So, C = 510.
After one half-life (38 days), the remaining amount will be 255. Therefore:
Solving for k:
Thus, our function is:
Then after one year or 365 days, the amount remaining will be about: