About 1313% of the population of a large country is hopelessly romantichopelessly romantic. If two people are randomly selecte
1 answer:
Answer:
Probability both are hopelessly romantic = 0.0169
Probability at least one is hopelessly romantic = 0.2431
Step-by-step explanation:
We are given that about 1313% of the population of a large country is hopelessly romantic.
Also, given two people are randomly selected.
Firstly, Let <em>Probability that the population of a large country is hopelessly romantic = P(R) = 0.13 </em>
(a) Probability both are hopelessly romantic is given by = Probability that first person is hopelessly romantic Probability that second person is hopelessly romantic
=
= 0.13 0.13 = <u>0.0169</u>
(b) Now, Probability that the population of a large country is not hopelessly romantic = P(R') = 1 - P(R) = 1 - 0.13 = 0.87.
So, Probability at least one is hopelessly romantic is given by = 1 - Probability that none of them is hopelessly romantic
=
= 1 - (0.87 0.87)
= 1 - 0.7569 = <u>0.2431</u>
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Answer:
Step-by-step explanation:
Six times a number is decreased by fourteen, the result is 124.
Write an equation that represents this
(6 * x) - 14 = 124
Multiply 6 by x to remove the parenthesis
6x - 14 = 124
Add 14 to both sides of the equation
6x = 138
Divide both sides of the equation by 6
x = 23
The unknown number is
Hope this helps :)
Answer:
P(X ≤ 46 | X~B(821, 0.078)) = 0.00885745584
0.00885... < 0.01
The test statistic of 46 is significant
There is sufficient evidence to reject H₀ and accept H₁
Air bags are more effective as protection than safety belts
Step-by-step explanation:
821 crashes
46 hospitalisations where car has air bags
7.8% or 0.078 probability of hospitalisations in cars with automatic safety belts
α = 0.01 or 1% ← level of significance
One-tailed test
We are testing whether hospitalisations in cars with air bags are less likely than in a car with automatic safety belts;
The likelihood of hospitalisation in a car with automatic safety belts, we are told, is 7.8% or 0.078;
So we are testing if hospitalisations in cars with air bags is less than 0.078;
So, firstly:
Let X be the continuous random variable, the number of hospitalisations from a car crash with equipped air bags
X~B(821, 0.078)
Null hypothesis (H₀): p = 0.078
Alternative hypothesis (H₁): p < 0.078
According to the information, we reject H₀ if:
P(X ≤ 46 | X~B(821, 0.078)) < 0.01
To find P(X ≤ 46) or equally P(X < 47), it could be quite long-winded to do manually for this particular scenario;
If you are interested, the manual process involves using the formula for every value of x up to and including 46, i.e. x = 0, x = 1, x = 2, etc. until x = 46, the formula is:
You can find binomial distribution calculators online, where you input n (i.e. the number of trials or 821 in this case), probability (i.e. 0.078) and the test statistic (i.e. 46), it does it all for you, which gives:
P(X ≤ 46 | X~B(821, 0.078)) = 0.00885745584
Now, we need to consider if the condition for rejecting H₀ is met and recognise that:
0.00885... < 0.01
There is sufficient evidence to reject H₀ and accept H₁.
To explain what this means:
The test statistic of 46 is significant according to the 1% significance level, meaning the likelihood that only 46 hospitalisations are seen in car crashes with air bags in the car as compared to the expected number in car crashes with automatic safety belts is very unlikely, less than 1%, to be simply down to chance;
In other words, there is 99%+ probability that the lower number of hospitalisations in car crashes with air bags is due to some reason, such as air bags being more effective as a protective implement than the safety belts in car crashes.