Question

Consider the probability that fewer than 26 out of 107 cell phone calls will be disconnected. Assume the probability that a given cell phone call will be disconnected is 98%. Specify whether the normal curve can be used as an approximation to the binomial probability by verifying the necessary conditions.

Answer 1

Answer:

We assume the condition of independence satisifed.

We need to check the following conditions in order to use the normal approximation.

$np=107*0.98=104.86 \geq 10$

$n(1-p)=107*(1-0.98)=2.14 < 10$

So we see that we satisfy the first condition, but the second no so then we can't apply the approximation for this case, since we need both conditions at the same time in order to use the normal approximation.

Step-by-step explanation:

Previous concepts

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".

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=107, p=0.98)$

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:

$nCx=\frac{n!}{(n-x)! x!}$

Solution to the problem

We assume the condition of independence satisifed.

We need to check the following conditions in order to use the normal approximation.

$np=107*0.98=104.86 \geq 10$

$n(1-p)=107*(1-0.98)=2.14 < 10$

So we see that we satisfy the first condition, but the second no so then we can't apply the approximation for this case, since we need both conditions at the same time in order to use the normal approximation.