Question

According to recent surveys, 60% of households have personal computers. If a random sample of 182 households is selected, what is the probability that more than 68 but fewer than 91 have a personal computer? Round z-value calculations to 2 decimal places and final answer to at least 4 decimal places.

Answer 1

The question involves calculating the probability of a given range of successes in a binomial distribution using normal approximation. Z-scores need to be computed for the values 68 and 91, then the cumulative probabilities are found and subtracted to obtain the final probability.

Step-by-step explanation:

The question asks for the probability that in a random sample of 182 households, more than 68 but fewer than 91 have a personal computer, given that 60% of households have personal computers. To solve this, we need to use the normal approximation to the binomial distribution since the sample size is large enough. The mean (μ) can be calculated as n × p, and the standard deviation (σ) as √n × p × (1-p), where n is the sample size and p is the probability of success.

To find the probabilities associated with the specific numbers of households, we convert the absolute numbers into z-scores using the formula: z = (X - μ) / σ. We then use these z-scores to look up the corresponding probabilities in the standard normal distribution table or use a calculator with the normalCDF function. Finally, to find the probability that more than 68 but fewer than 91 households have a personal computer, we subtract the cumulative probability up to 68 from the cumulative probability up to 91.

Answer 2

To answer this question we are going to perform a normal approximation to the binomial distribution.

We know that if X is binomial random variable we can approximate it to a normal distribution with:

`\begin{gathered} \mu=np \\ \text{and} \\ \sigma=\sqrt[]{npq} \end{gathered}`

In this case we have that:

`\begin{gathered} p=0.6 \\ q=1-p=1-0.6=0.4 \\ n=182 \end{gathered}`

Then the approximate normal distribution will have mean and standard deviation:

`\begin{gathered} \mu=(0.6)(182)=109.2 \\ \sigma=\sqrt[]{(0.6)(0.4)(182)}=6.61 \end{gathered}`

Now that we have the normal approximate distribution we need to calculate for this normal distribution:

we have that:

[tex]\begin{gathered} P(68Using a normal distribution table:[tex]\begin{gathered} P(68Therefore the probability we are looking for is 0.00298