Answer:
0.0485 = 4.85% probability that you miss the bus.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean 
and standard deviation 
, the z-score of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
When two normal distributions are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.
In this question:
We have to find the distribution for the difference in times between when you arrive and when the bus arrives.
You arrive at 8, so we consider the mean 0. The bus arrives at 8:05, 5 minutes later, so we consider mean 5. This means that the mean is:
The standard deviation of your arrival time is of 2 minutes, while for the bus it is 3. So
The bus remains at the stop for 1 minute and then leaves. What is the chance that I miss the bus?
You will miss the bus if the difference is larger than 1. So this probability is 1 subtracted by the pvalue of Z when X = 1.
has a pvalue of 0.9515
1 - 0.9515 = 0.0485
0.0485 = 4.85% probability that you miss the bus.
Again u just have to meaiuse
1) Our marbles will be blue, red, and green. You need two fractions that can be multiplied together to make 1/6. There are two sets of numbers that can be multiplied to make 6: 1 and 6, and 2 and 3. If you give the marbles a 1/1 chance of being picked, then there's no way that a 1/6 chance can be present So we need to use a 1/3 and a 1/2 chance. 2 isn't a factor of 6, but 3 is. So we need the 1/3 chance to become apparent first. Therefore, 3 of the marbles will need to be one colour, to make a 1/3 chance of picking them out of the 9. So let's say 3 of the marbles are green. So now you have 8 marbles left, and you need a 1/2 chance of picking another colour. 8/2 = 4, so 4 of the marbles must be another colour, to make a 1/2 chance of picking them. So let's say 4 of the marbles are blue. We know 3 are green and 4 are blue, 3 + 4 is 7, so the last 2 must be red.
The problem could look like this:
A bag contains 4 blue marbles, 2 red marbles, and 3 green marbles. What are the chances she will pick 1 blue and 1 green marble?
You should note that picking the blue first, then the green, will make no difference to the overall probability, it's still 1/6. Don't worry, I checked
2) a - 2% as a probability is 2/100, or 1/50. The chance of two pudding cups, as the two aren't related, both being defective in the same packet are therefore 1/50 * 1/50, or 1/2500.
b - 1,000,000/2500 = 400
400 packages are defective each year
1. 7.5 because Y/75 X 10/100 = 750 /100 = 7.5
2. .75 because Y/75 X 1/100 = 75/100 = .75
I believe they compare because both butterflies can have a wingspan of 3.5 in. Hope this helps
