A survey was conducted to measure the number of hours per week adults spend on home
computers. In the survey, the number of hours was normally distributed, with a mean of 8
hours and a standard deviation of 1 hour. A survey participant is randomly selected. Find the
probability that the hours spent on the home computer by the participant are between 5.5 and
9.5 hours per week.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to your Z scores.
0.3258
Z-scores: (9.5-8)/1=1.5 and (5.5-8)/1=-2.5. Corresponding probabilities: 94.41%
and 0.62%, so the answer is P=94.41%-0.62%=93.79%
To find the probability that the hours spent on the home computer by the participant are between 5.5 and 9.5 hours per week, we need to calculate the z-scores for the lower and upper limits and then find the area under the normal curve between those z-scores.
The z-score formula is: z = (x - μ) / σ
where:
z = z-score
x = value we're interested in (in this case, the lower and upper limits of hours per week)
μ = mean of the distribution (8 hours per week)
σ = standard deviation (1 hour per week)
For the lower limit of 5.5 hours:
z_lower = (5.5 - 8) / 1 = -2.5
For the upper limit of 9.5 hours:
z_upper = (9.5 - 8) / 1 = 1.5
Now, we need to find the probability of the area between these z-scores. We can use a standard normal distribution table or a calculator with a normal distribution function to find these probabilities.
Using a standard normal distribution table, we can look up the probabilities associated with the z-scores -2.5 and 1.5. The table will give us the area to the left of each z-score.
From the table, we find the following probabilities:
P(z < -2.5) = 0.0062 (approximately)
P(z < 1.5) = 0.9332 (approximately)
To find the probability of the area between -2.5 and 1.5, we subtract the smaller probability from the larger probability:
P(-2.5 < z < 1.5) = P(z < 1.5) - P(z < -2.5) = 0.9332 - 0.0062 = 0.927 (approximately)
Therefore, the probability that the hours spent on the home computer by the participant are between 5.5 and 9.5 hours per week is approximately 0.927, or 92.7%.