Use the appropriate formula (binompdf(n, p, x) or binomcdf(n, p, x)) to solve problems (a) and (b). The National Health Interview Survey reports that 40% of telephone users no longer use landlines and have switched completely to cell phone use. Suppose we take a random sample of 10 telephone users. a. Find the probability that the sample contains two users who have abandoned their landlines. (First identify n, p, x, and use the correct function. Show work). b. Find the probability that the sample contains at most two users who have abandoned their landlines (first identify n, p, x, and then use the correct function. Show work). c. Use the Binomial probability formula to answer question (a) above. P(x) = C(n, x) * p^x * q^(n-x) = n! / (x!(n-x)!) * (p^x) * (q^(n-x))
The Correct Answer and Explanation is:
Let’s break down the problem step by step.
Part (a) Find the probability that the sample contains two users who have abandoned their landlines.
Given:
- Probability of a user abandoning landlines, p=0.40p = 0.40p=0.40.
- Probability of a user not abandoning landlines, q=1−p=0.60q = 1 – p = 0.60q=1−p=0.60.
- The sample size is n=10n = 10n=10.
- We want to find the probability that exactly two users have abandoned their landlines, so x=2x = 2x=2.
This is a binomial probability problem, and we can use the binompdf function (binomial probability mass function).
The formula for binomial probability is:P(x)=(nx)⋅px⋅qn−xP(x) = \binom{n}{x} \cdot p^x \cdot q^{n – x}P(x)=(xn)⋅px⋅qn−x
where:
- (nx)=n!x!(n−x)!\binom{n}{x} = \frac{n!}{x!(n – x)!}(xn)=x!(n−x)!n! is the binomial coefficient.
Substitute the given values into the formula:P(2)=(102)⋅(0.40)2⋅(0.60)8P(2) = \binom{10}{2} \cdot (0.40)^2 \cdot (0.60)^8P(2)=(210)⋅(0.40)2⋅(0.60)8
Now calculate:
- (102)=10!2!(10−2)!=10×92×1=45\binom{10}{2} = \frac{10!}{2!(10 – 2)!} = \frac{10 \times 9}{2 \times 1} = 45(210)=2!(10−2)!10!=2×110×9=45
- (0.40)2=0.16(0.40)^2 = 0.16(0.40)2=0.16
- (0.60)8≈0.0168(0.60)^8 \approx 0.0168(0.60)8≈0.0168
So,P(2)=45×0.16×0.0168≈0.12096P(2) = 45 \times 0.16 \times 0.0168 \approx 0.12096P(2)=45×0.16×0.0168≈0.12096
Therefore, the probability that exactly two users have abandoned their landlines is approximately 0.12096, or about 12.1%.
Part (b) Find the probability that the sample contains at most two users who have abandoned their landlines.
For this, we need to calculate the probability of having 0, 1, or 2 users who have abandoned their landlines, and then sum these probabilities.
We use the binomcdf function (binomial cumulative distribution function) for this. The function gives the probability of having x or fewer successes.
We will find P(X≤2)P(X \leq 2)P(X≤2).
This can be written as:P(X≤2)=P(0)+P(1)+P(2)P(X \leq 2) = P(0) + P(1) + P(2)P(X≤2)=P(0)+P(1)+P(2)
We already found P(2)P(2)P(2) above. Now, let’s calculate P(0)P(0)P(0) and P(1)P(1)P(1):
- P(0)=(100)⋅(0.40)0⋅(0.60)10≈1×1×0.0060=0.0060P(0) = \binom{10}{0} \cdot (0.40)^0 \cdot (0.60)^{10} \approx 1 \times 1 \times 0.0060 = 0.0060P(0)=(010)⋅(0.40)0⋅(0.60)10≈1×1×0.0060=0.0060
- P(1)=(101)⋅(0.40)1⋅(0.60)9≈10×0.40×0.0101=0.0404P(1) = \binom{10}{1} \cdot (0.40)^1 \cdot (0.60)^9 \approx 10 \times 0.40 \times 0.0101 = 0.0404P(1)=(110)⋅(0.40)1⋅(0.60)9≈10×0.40×0.0101=0.0404
Thus,P(X≤2)=P(0)+P(1)+P(2)=0.0060+0.0404+0.12096=0.16736P(X \leq 2) = P(0) + P(1) + P(2) = 0.0060 + 0.0404 + 0.12096 = 0.16736P(X≤2)=P(0)+P(1)+P(2)=0.0060+0.0404+0.12096=0.16736
So, the probability that the sample contains at most two users who have abandoned their landlines is approximately 0.16736, or about 16.7%.
Part (c) Use the Binomial probability formula to answer question (a) above.
We already solved part (a) using the binomial probability formula:P(2)=(102)⋅(0.40)2⋅(0.60)8P(2) = \binom{10}{2} \cdot (0.40)^2 \cdot (0.60)^8P(2)=(210)⋅(0.40)2⋅(0.60)8
and the result was approximately 0.12096, as calculated earlier.
Summary of Results:
- (a) The probability that the sample contains exactly two users who have abandoned their landlines is approximately 0.12096 (12.1%).
- (b) The probability that the sample contains at most two users who have abandoned their landlines is approximately 0.16736 (16.7%).
This problem is based on the binomial distribution, which models situations where there are exactly two outcomes (success or failure) across a fixed number of trials, such as whether a user abandons their landline. The formulas used were the binomial probability mass function and cumulative distribution function.
