Answer:
a.) 0.7063
b.) 23
Step-by-step explanation:
a.)
Let X be an event in which at least 2 students have same birthday
Y be an event in which no student have same birthday.
Now,
P(X) + P(Y) = 1
⇒P(X) = 1 - P(Y)
as we know that,
Probability of no one has birthday on same day = P(Y)
⇒P(Y) =
where there are n people in a group
As given,
n = 30
⇒P(Y) =
= 0.2937
∴ we get
P(X) = 1 - 0.2937 = 0.7063
So,
The probability that at least two of them have their birthdays on the same day = 0.7063
b.)
Given, P(X) > 0.5
As
P(X) + P(Y) = 1
⇒P(Y) ≤ 0.5
As
P(Y) =
We use hit and trial method
If n = 1 , then
P(Y) =
= 1
0.5
If n = 5 , then
P(Y) =
= 0.97
0.5
If n = 10 , then
P(Y) =
= 0.88
0.5
If n = 15 , then
P(Y) =
= 0.75
0.5
If n = 20 , then
P(Y) =
= 0.588
0.5
If n = 22 , then
P(Y) =
= 0.52
0.5
If n = 23 , then
P(Y) =
= 0.49
0.5
∴ we get
Number of students should be in class in order to have this probability above 0.5 = 23
Answer:
-13-4i
Step-by-step explanation:
This answer is correct or not please tell me
Answer:
In mathematics, the empty set is the unique set having no elements; its size or cardinality (count of elements in a set) is zero.
Hope this helps!
Step-by-step explanation:
Answer: A) .1587
Step-by-step explanation:
Given : The amount of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce.
i.e.
and 
Let x denotes the amount of soda in any can.
Every can that has more than 12.50 ounces of soda poured into it must go through a special cleaning process before it can be sold.
Then, the probability that a randomly selected can will need to go through the mentioned process = probability that a randomly selected can has more than 12.50 ounces of soda poured into it =
![P(x>12.50)=1-P(x\leq12.50)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{12.50-12.30}{0.20})\\\\=1-P(z\leq1)\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.8413\ \ \ [\text{By z-table}]\\\\=0.1587](https://tex.z-dn.net/?f=P%28x%3E12.50%29%3D1-P%28x%5Cleq12.50%29%5C%5C%5C%5C%3D1-P%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5Cleq%5Cdfrac%7B12.50-12.30%7D%7B0.20%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq1%29%5C%20%5C%20%5B%5Cbecause%20z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3D1-0.8413%5C%20%5C%20%5C%20%5B%5Ctext%7BBy%20z-table%7D%5D%5C%5C%5C%5C%3D0.1587)
Hence, the required probability= A) 0.1587