Question

Consider the probability that at least 88 out of 158 software users will not call technical support. Assume the probability that a given software user will not call technical support is 57%. Approximate the probability using the normal distribution. Round your answer to four decimal places.

Answer 1

Answer:

$E(X) = 158*0.57 =90.06$

And the standard deviation for the random variable is given by:

$\sigma= \sqrt{158*0.57*(1-0.57)} = 6.223$

And the distribution for the approximation is given by:

$X \sim N (\mu = 90.06, \sigma = 6.223)$

We can use the z score formula given by:

$z = \frac{x -\mu}{\sigma}$

And using the complement rule we got:

$P(X \geq 88) = 1-P(X$

And using the standard normal table we got:

$P(X \geq 88) = 1-P(X$

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".  

Solution to the problem

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

$X \sim Bin (n = 158, p =0.57)$

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:  

$nCx=\frac{n!}{(n-x)! x!}$  

We want this probability:

$P(X\geq 88)$

Normal approximation

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

$np =158*0.57 = 90.06>10$

$n(1-p) = 158*(1-0.57) = 67.94>10$

Since we satisfy both conditions the normal approximation makes sense

The expected value is given by this formula:

$E(X) = 158*0.57 =90.06$

And the standard deviation for the random variable is given by:

$\sigma= \sqrt{158*0.57*(1-0.57)} = 6.223$

And the distribution for the approximation is given by:

$X \sim N (\mu = 90.06, \sigma = 6.223)$

We can use the z score formula given by:

$z = \frac{x -\mu}{\sigma}$

And using the complement rule we got:

$P(X \geq 88) = 1-P(X$

And using the standard normal table we got:

$P(X \geq 88) = 1-P(X$