Mika used division to rewrite 5/40
2 answers:
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Answer:
0.62% probability that a random sample of 16 bulbs will have an average life of less than 775 hours.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
Normal probability distribution.
Problems of normally distributed samples can be solved using 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 problem, we have that:
Find the probability that a random sample of 16 bulbs will have an average life of less than 775 hours.
This probability is the pvalue of Z when . So
has a pvalue of 0.0062. So there is a 0.62% probability that a random sample of 16 bulbs will have an average life of less than 775 hours.
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