You can model that you expect a 1.25% raise each year that you work for a certain company. If you currently make $40,000, how ma
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Answer:
a) 2.84% probability that he is late for his first lecture.
b) 5.112 days
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean and standard deviation
, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
a. Find the probability that he is late for his first lecture.
This is the probability that he takes more than 20 minutes to walk, which is 1 subtracted by the pvalue of Z when X = 20. So
has a pvalue of 0.9716
1 - 0.9716 = 0.0284
2.84% probability that he is late for his first lecture.
b. Find the number of days per year he is likely to be late for his first lecture.
Each day, 2.84% probability that he is late for his first lecture.
Out of 180
0.0284*180 = 5.112 days
Answer:
Probability that Adam and Ella will be selected:
Step-by-step explanation:
Probabilities
The probability of a random event E to occur is a real number between 0 and 1, both inclusive, where 0 indicates an impossible event and 1 a sure event. There are many techniques to compute probabilities depending on the particular situation and distribution.
This question will be solved by simple calculations and logic, given its simplicity. We know the middle school chess club has 5 members: Adam, Bradley, Carol, Dave, and Ella. Two of them are going to be selected at random to participate in the county chess tournament. We can calculate the number of different ways it can be done without any restriction. It's called the sample space.
The sample space of this event is the combination of 5 members regardless of their position. If {a,b,c,d,e} are the five members, then the possible combinations are {ab,ac,ad,ae,bc,bd,be,cd,ce,de}. Notice that there are only 10 possibilities because the combination ab is the same as ba since it's the same team for the tournament.
We can see there is only one combination of two specific letters out of 10. If a=Adam and e=Ella, only one combination is ae, the other 9 don't include both members, so the probability is