. In his management information systems textbook, Professor David Kroenke raises an interesting point: “If 98% of our market has Internet access, do we have a responsibility to provide non-Internet materials to that other 2%? Suppose that 98% of the customers in your market do have Internet access, and you select a random sample of 500 customers with a standard error of the mean of .006261. What is the probability that the sample has
a. Greater than 99% of the customers with internet access?
b. Between 97% and 99% of the customers with Internet access?
c. Fewer than 97% of the customers with Internet access?
To find the probabilities for each scenario, we need to use the standard error of the mean and calculate the z-scores. The z-score measures how many standard deviations a given value is away from the mean.
First, let's calculate the mean and standard deviation based on the given information. We know that the sample size is 500, and the standard error of the mean is 0.006261. The standard deviation would be the standard error of the mean multiplied by the square root of the sample size.
Mean = 98% = 0.98
Standard Error = 0.006261
Sample Size = 500
Standard Deviation = Standard Error * sqrt(Sample Size)
= 0.006261 * sqrt(500)
≈ 0.279
Now, we can calculate the z-scores for each scenario.
a. Greater than 99% of the customers with internet access:
To find the probability that the sample has greater than 99% of the customers with internet access, we need to find the z-score corresponding to 99%. The formula for z-score is (X - Mean) / Standard Deviation.
Z-score = (99% - Mean) / Standard Deviation
= (0.99 - 0.98) / 0.279
≈ 0.036
Using a standard normal distribution table or a calculator, we can find the probability associated with this z-score. Subtracting this probability from 1 will give us the desired probability.
b. Between 97% and 99% of the customers with Internet access:
To find the probability that the sample has between 97% and 99% of the customers with internet access, we need to find the z-scores corresponding to both percentages.
Z1 = (97% - Mean) / Standard Deviation
= (0.97 - 0.98) / 0.279
≈ -0.036
Z2 = (99% - Mean) / Standard Deviation
= (0.99 - 0.98) / 0.279
≈ 0.036
We can then find the probability associated with each z-score and subtract Z1 from Z2 to get the desired probability.
c. Fewer than 97% of the customers with Internet access:
To find the probability that the sample has fewer than 97% of the customers with internet access, we can find the z-score corresponding to 97%.
Z-score = (97% - Mean) / Standard Deviation
= (0.97 - 0.98) / 0.279
≈ -0.036
Using a standard normal distribution table or a calculator, we can find the probability associated with this z-score.
Please note that the final probabilities will depend on the specific values obtained from the z-score calculations and the probability tables used.
To answer these probability questions, we need to use the normal distribution and the concept of standard error. The standard error of the mean allows us to estimate the variability in our sample mean compared to the population mean.
To calculate the probabilities, we will use the Z-score formula, which indicates how many standard deviations a particular value is from the mean. The formula for the Z-score is:
Z = (X - μ) / σ
Where:
- X is the value we are interested in (e.g., percentage of customers with internet access)
- μ is the mean of the population (98% in this case)
- σ is the standard deviation of the population (calculated using the standard error of the mean)
We can then use the Z-score to look up the probabilities using a standard normal distribution table or a statistical calculator.
a. Probability of having greater than 99% of customers with internet access:
To calculate this probability, we need to find the Z-score of 99% and use the area under the normal distribution curve to the right of that Z-score.
Z = (99 - 98) / 0.006261
Z ≈ 159.58
Using a standard normal distribution table or calculator, we can find the probability associated with a Z-score of 159.58. This probability is extremely close to 1 or 100%, since it represents an extremely large deviation from the mean. Hence, the probability of having greater than 99% of customers with internet access is practically 100%.
b. Probability of having between 97% and 99% of customers with internet access:
To calculate this probability, we need to find the Z-scores of 97% and 99% and use the difference between their areas under the normal distribution curve.
Z1 = (97 - 98) / 0.006261 ≈ -159.58
Z2 = (99 - 98) / 0.006261 ≈ 159.58
Using a standard normal distribution table or calculator, we can find the probabilities associated with these Z-scores and subtract them to find the desired probability. Since both Z-scores are extremely large (one negative and one positive), their probabilities are practically 0. Therefore, the probability of having between 97% and 99% of customers with internet access is practically 0%.
c. Probability of having fewer than 97% of customers with internet access:
To calculate this probability, we need to find the Z-score of 97% and use the area under the normal distribution curve to the left of that Z-score.
Z = (97 - 98) / 0.006261 ≈ -159.58
Using a standard normal distribution table or calculator, we can find the probability associated with a Z-score of -159.58. This probability is also extremely close to 0, since it represents an extremely large deviation from the mean in the opposite direction. Hence, the probability of having fewer than 97% of customers with internet access is practically 0%.
In conclusion:
a. The probability of having greater than 99% of customers with internet access is practically 100%.
b. The probability of having between 97% and 99% of customers with internet access is practically 0%.
c. The probability of having fewer than 97% of customers with internet access is practically 0%.