The time it takes Jessica to bicycle to school is normally distributed with an average of a third of an hour and a variance of 9 minutes. Jessica has to be at school at 8:15 am. What time should she leave the house so she will be late on 5% of the time?
Z = (score-mean)/SD
1/3 hour = 20 minutes
Variance = SD^2
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion .05 and insert its Z score in the equation above with the other data. Find difference from 8:15.
To find the time Jessica should leave the house so she will be late on 5% of the time, we need to find the z-score corresponding to the 5th percentile of the normal distribution and convert it back to the time scale.
Step 1: Convert variance to standard deviation
Since variance is given in minutes, we need to convert it to standard deviation by taking the square root. The variance is 9 minutes, so the standard deviation (σ) is √9 = 3 minutes.
Step 2: Find the z-score
The z-score is a measure of how many standard deviations an observation is from the mean. To find the z-score corresponding to the 5th percentile, we need to look up the z-score from the standard normal distribution table. The 5th percentile corresponds to a z-score of -1.645 (rounded to three decimal places).
Step 3: Convert z-score back to time
To convert the z-score back to the time scale, we multiply it by the standard deviation and add it to the mean.
Late time = Average time + (z-score * standard deviation)
= 8:15 am + (-1.645 * 3 minutes)
= 8:15 am - 4.935 minutes
Step 4: Convert minutes to hours
To convert the negative minutes to hours, we need to subtract the fractional part from 60 and consider the whole number part as an hour.
Late time = 8:15 am - 4.935 minutes
= 8:15 am - 4 minutes and 56.1 seconds
= 8:15 am - 4 minutes and 1 second
= 8:11 am and 59 seconds
Therefore, Jessica should leave the house around 8:11 am and 59 seconds to be late on 5% of the time.