<h3>
Answer: 34%</h3>
This result is approximate.
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Explanation:
mu = 750 = mean
sigma = 75 = standard deviation
The raw scores or x values are x = 750 and x = 825
Let's compute the z score for each x value
z = (x - mu)/sigma
z = (750 - 750)/75
z = 0
and
z = (x - mu)/sigma
z = (825 - 750)/75
z = 1
Therefore P(750 ≤ x ≤ 825) is equivalent to P(0 ≤ z ≤ 1) in this context.
Use a z score table to determine that
P(z ≤ 0) = 0.5
P(z ≤ 1) = 0.84314 approximately
So,
P(a ≤ z ≤ b) = P(z ≤ b) - P(z ≤ a)
P(0 ≤ z ≤ 1) = P(z ≤ 1) - P(z ≤ 0)
P(0 ≤ z ≤ 1) = 0.84314 - 0.5
P(0 ≤ z ≤ 1) = 0.34314 approximately
The value 0.34314 then converts to 34.314% which rounds to <u>34%</u>
Or you could use the empirical rule as shown below. The pink section on the right is marked <u>34%</u> which is approximate. This pink section is between z = 0 and z = 1.
You’re finding the difference between the two numbers (subtraction). 1/4 is 0.25.
40.25 - 38.75 = 1.5
She worked one and a half more hours in the second week.
Answer:
5
Step-by-step explanation:
12+12+12= 36
36-7= 29
3+3+3=9
29-9=20
20÷4=5
Answer:
Correct option: (a) 0.1452
Step-by-step explanation:
The new test designed for detecting TB is being analysed.
Denote the events as follows:
<em>D</em> = a person has the disease
<em>X</em> = the test is positive.
The information provided is:
Compute the probability that a person does not have the disease as follows:
The probability of a person not having the disease is 0.12.
Compute the probability that a randomly selected person is tested negative but does have the disease as follows:
![P(X^{c}\cap D)=P(X^{c}|D)P(D)\\=[1-P(X|D)]\times P(D)\\=[1-0.97]\times 0.88\\=0.03\times 0.88\\=0.0264](https://tex.z-dn.net/?f=P%28X%5E%7Bc%7D%5Ccap%20D%29%3DP%28X%5E%7Bc%7D%7CD%29P%28D%29%5C%5C%3D%5B1-P%28X%7CD%29%5D%5Ctimes%20P%28D%29%5C%5C%3D%5B1-0.97%5D%5Ctimes%200.88%5C%5C%3D0.03%5Ctimes%200.88%5C%5C%3D0.0264)
Compute the probability that a randomly selected person is tested negative but does not have the disease as follows:
![P(X^{c}\cap D^{c})=P(X^{c}|D^{c})P(D^{c})\\=[1-P(X|D)]\times{1- P(D)]\\=0.99\times 0.12\\=0.1188](https://tex.z-dn.net/?f=P%28X%5E%7Bc%7D%5Ccap%20D%5E%7Bc%7D%29%3DP%28X%5E%7Bc%7D%7CD%5E%7Bc%7D%29P%28D%5E%7Bc%7D%29%5C%5C%3D%5B1-P%28X%7CD%29%5D%5Ctimes%7B1-%20P%28D%29%5D%5C%5C%3D0.99%5Ctimes%200.12%5C%5C%3D0.1188)
Compute the probability that a randomly selected person is tested negative as follows:
Thus, the probability of the test indicating that the person does not have the disease is 0.1452.
