Suppose that 92% of all SFSU students own a cell phone. Find the probability that 7 out of 10 random selected SFSU students own a cell phone.
explain how answer was arrived
P(7) = (0.92)^7*(0.08)^3*[10!/(7!*3!)]
= (10*9*8)/(1*2*3)]*(0.92)^7*(0.08)^3
= 0.03427
P(10)= 0.92^10 = 0.43439
P(9) = 0.92^9*0.08*10 = 0.37777
P(8) = 0.92^8*0.08^2*(10*9/2)=0.14781
To find the probability that 7 out of 10 randomly selected SFSU students own a cell phone, we can use the binomial probability formula.
The binomial probability formula is given by:
P(x) = nCx * p^x * (1-p)^(n-x)
Where:
- P(x) is the probability of getting exactly x successes
- n is the total number of trials (in this case, the number of students selected)
- x is the number of successes (in this case, the number of students who own a cell phone)
- p is the probability of success (in this case, the proportion of students who own a cell phone)
- nCx represents the number of ways to choose x successes out of n trials, and can be calculated using combinations.
In this case, we are given that 92% of all SFSU students own a cell phone, so p = 0.92. We want to find the probability of exactly 7 students out of 10 owning a cell phone, so n = 10 and x = 7.
Now, let's calculate it step by step:
1. Calculate nCx:
nCx = n! / (x!(n-x)!), where "!" represents factorial notation.
In this case,
10C7 = 10! / (7!(10-7)!) = 10! / (7! * 3!) = (10 * 9 * 8) / (3 * 2 * 1) = 120.
2. Calculate the probability of success raised to the power of x:
p^x = 0.92^7 = 0.4783.
3. Calculate the probability of failure raised to the power of (n - x):
(1-p)^(n-x) = (1-0.92)^(10-7) = 0.0776.
4. Multiply all the calculated values together:
P(7 out of 10 students own a cell phone) = 120 * 0.4783 * 0.0776 = 0.275.
Therefore, the probability that 7 out of 10 randomly selected SFSU students own a cell phone is approximately 0.275, or 27.5%.