The average time to sew a pair of pants is 4.2 hours with a standard deviation of 30 minutes and if the distribution is normal then the probability of a worker finishing the pants in less than 3 hours is.
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 that Z score.
The unemployment rate in a city is %. Find the probability that fewer than out of people from this city sampled at random are unemployed. Round your answer to four decimal places.
To find the probability of a worker finishing the pants in less than 3 hours, we need to standardize the value using the z-score formula and then find the corresponding area under the standard normal distribution curve.
First, we need to calculate the z-score using the given information. The formula for the z-score is:
z = (x - μ) / σ
Where:
x = the value we want to find the probability for (in this case, 3 hours)
μ = mean of the distribution (4.2 hours)
σ = standard deviation of the distribution (30 minutes, or 0.5 hours)
Plugging in the values:
z = (3 - 4.2) / 0.5
z = -1.2 / 0.5
z = -2.4
Now, we need to find the area to the left of this z-score in the standard normal distribution table. This represents the probability of a worker finishing the pants in less than 3 hours.
Using a standard normal distribution table or a statistical calculator, we find that the area to the left of -2.4 is approximately 0.0082.
Therefore, the probability of a worker finishing the pants in less than 3 hours is approximately 0.0082, or 0.82%.