si franco comió 8/3 de pizza y Fabián comió 5/6 de la misma pizza. ¿quien comió más ? si quedó 4/9 de pizza.
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Answer:
17.80% probability that all of them are wearing their seat belts.
Step-by-step explanation:
For each driver stopped, there are only two possible outcomes. Either they are wearing their seatbelts, or they are not. The drivers are chosen at random, which mean that the probability of a driver wearing their seatbelts is independent from other drivers. So we use the normal probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
Police estimate that 25% of drivers drive without their seat belts.
This means that 75% wear their seatbelts, so
If they stop 6 drivers at random, find the probability that all of them are wearing their seat belts.
This is .
17.80% probability that all of them are wearing their seat belts.
Answer:
C.-2.33
Step-by-step explanation:
We are given that
Significance level, %=0.01
We have to find the critical value of z for a left-tailed test.
To find the critical value , we will use the z-table at significance level 0.01 for left-tailed test.
By using the z-table at significance level 0.01 for left-tailed test
The critical value of z at 1% significance level for left-tailed test
=-2.33
We get p-value by using z-critical value
P-value=0.0099<0.01
So, hypothesis test would be reject the null hypothesis at at 1% significance level.
Hence, the critical value of z=-2.33
Option C is true.