Answer:
A) 12.29%
B) 81.33%
C) 1.55%
Explanation:
This is a binomial model, and I assume you're meant to use a calculator to solve.
A)
This is what I have in my notes for the binompdf function:
- binompdf(n, p, x) - find the probability of exactly x successes in n trials given p probability
That is exactly what we want in this case, we're looking for exactly 5 people who meet the conditions (successes) given n trials and p probability.
Using a calculator:
binompdf(10, 0.68, 5) ≈ 0.1229
That means you have about a 12.29% chance to randomly select exactly 5 adults who have little confidence in newspapers.
B)
For this one, use the binomcdf function:
- binomcdf(n, p, x) - find the probability of x or fewer successes in n trials given p probability
In this case, we're looking for more, not less. However, you can think of the probability of getting more than 6 as just the probability of not getting less than 6.
With a calculator:
binomcdf(10, 0.68, 5) ≈ 0.1867
That means you have about an 18.67% chance to randomly select 5 or less people. We're looking for more though. As said above, the probability of getting 6 or more is just the probability of not getting 5 or less. To find the probability of something not happening, just subtract the probability from 1.
1 - 0.1867 = 0.8133
An 81.33% chance to randomly select at least 6 adults who have little confidence in newspapers.
C)
Same as the above, I don't see a need to explain it all again. Use binomcdf:
binomcdf(10, 0.68, 3) = 0.0155
Notice I used 3 rather than 4. binomcdf calculates x or fewer, or 'rather less than or equal to'. We want just 'less than 4', and that means 'less than or equal to 3'.
That's a 1.55% chance to select less than 4 adults with little confidence in newspapers.