Question

A CBS News/New York Times survey found that 97% of Americans believe that texting while driving should be outlawed (CBS News website, January 5, 2015).

a. For a sample of 10 Americans, what is the probability that at least 8 say that they believe texting while driving should be outlawed? Use the binomial distribution probability function to answer this question (to 4 decimals).

b. For a sample of 100 Americans, what is the probability that at least 95 say that they believe texting while driving should be outlawed? Use the normal approximation of the binomial distribution to answer this question. (to 4 decimals).

Answer 1

Answer:

a) $P(X\geq 8)=0.0317+0.228+0.737=0.9972$

b) $P(X>95) =1-P(X<95) = 1-P(Z<-1.17)=1-0.121=0.8790$

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

$P(X)=(nCx)(p)^x (1-p)^{n-x}$  

Where (nCx) means combinatory and it's given by this formula:  

$nCx=\frac{n!}{(n-x)! x!}$  

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: $P(A)+P(A') =1$

Part a

We want this probability:

$P(X\leq 12)$

$P(X\geq 8)=P(X=8)+P(X=9)+P(X=10)$

$P(X=8)=(10C8)(0.97)^8 (1-0.97)^{10-8}=0.03174$  

$P(X=9)=(10C9)(0.97)^9 (1-0.97)^{10-9}=0.2287$  

$P(X=10)=(10C10)(0.97)^{10} (1-0.97)^{10-10}=0.7374$  

$P(X\geq 8)=0.0317+0.228+0.737=0.9972$

Part b

We need to check if we can use the normal approximation , the conditions are:

$np=100*0.97=97>10$ and $n(1-p)=100*(1-0.97)=3$

On this case the second condition is not satisfied, but the problem says that we can use it. So then if we apply the normal approximation to the binomial distribution in our case:

$X \sim N(\mu=97,\sigma=1.706)$

We can use the z score formula given by:

$z=\frac{x-\mu}{\sigma}$  

And we want this probability:

$P(X>95) = P(Z>\frac{95-97}{1.706})= P(Z>-1.17)$

And we can use the complment rule and we got:

$P(X>95) =1-P(X<95) = 1-P(Z<-1.17)=1-0.121=0.8790$