Question

The speed at which you can log into a website through asmartphone is an important quality characteristic of that website.In a recent test, the mean time to log into the JetBlue Airwayswebsite through a smartphone was 4.237 seconds. (Data extractedfrom N. Trejos, "Travelers Have No Patience for Slow MobileSites," USA Today, April 4, 2012, p. 3B.) Suppose that the downloadtime is normally distributed, with a standard deviation of1.3 seconds. What is the probability that a download time is

a. less than 2 seconds?
b. between 1.5 and 2.5 seconds?
c. above 1.8 seconds?
d. Ninety-nine percent of the download times are slower (higher)than how many seconds?
e. Ninety-five percent of the download times are between whattwo values, symmetrically distributed around the mean?

Answer 1

The probability that a download time is less than 2 seconds is approximately 4.27%. The probability that a download time is between 1.5 and 2.5 seconds is approximately 10.9%. The probability that a download time is above 1.8 seconds is approximately 99.5%.

Step-by-step explanation:

To solve this problem, we will use the standard normal distribution with a mean of 0 and a standard deviation of 1. We can convert the download time to a standard score using the formula z = (x - mean) / standard deviation. Once we have the standard score, we can use a standard normal distribution table or calculator to find the probabilities.

a. To find the probability that the download time is less than 2 seconds, we need to find the area under the standard normal curve to the left of 2. Using a standard normal distribution table or calculator, we find that the z-score for 2 seconds is z = (2 - 4.237) / 1.3 = -1.72. The probability of a download time less than 2 seconds is approximately 0.0427, or 4.27%.

b. To find the probability that the download time is between 1.5 and 2.5 seconds, we need to find the area under the standard normal curve between the z-scores for 1.5 and 2.5 seconds. The z-score for 1.5 seconds is z = (1.5 - 4.237) / 1.3 = -2.10, and the z-score for 2.5 seconds is z = (2.5 - 4.237) / 1.3 = -1.33. Using a standard normal distribution table or calculator between these z-scores, we find that the probability is approximately 0.109, or 10.9%.

c. To find the probability that the download time is above 1.8 seconds, we need to find the area under the standard normal curve to the right of 1.8. The z-score for 1.8 seconds is z = (1.8 - 4.237) / 1.3 = -2.63. Using a standard normal distribution table or calculator, we find that the probability is approximately 0.995, or 99.5%.

d. To find the download time where 99% of the times are slower (higher), we need to find the z-score that corresponds to a probability of 1 - 0.99 = 0.01. Using a standard normal distribution table or calculator, we find that the z-score is approximately -2.33. We can solve for x by rearranging the standard score formula: x = mean + (z * standard deviation) = 4.237 + (-2.33 * 1.3) = 0.722 seconds.

e. To find the download times between which 95% of the times fall, we need to find the z-scores that correspond to a probability of 0.025 (for the lower bound) and a probability of 1 - 0.025 = 0.975 (for the upper bound). Using a standard normal distribution table or calculator, we find that the z-score for 0.025 is approximately -1.96, and the z-score for 0.975 is approximately 1.96. We can solve for x by rearranging the standard score formula: lower bound = mean + (z * standard deviation) = 4.237 + (-1.96 * 1.3) = 1.742 seconds, and upper bound = mean + (z * standard deviation) = 4.237 + (1.96 * 1.3) = 6.732 seconds.