Ryan won 60 tickets playing a game at the arcade. He gave four tickets
2 answers:
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Create a graph to locate the intersection of the equations. The intersection of the system of equations is the solution.
(18,269.82)
(18,269.82)
Answer:
(a) P(X 20) = 0.9319
(b) Expected number of defective light bulbs = 15
Step-by-step explanation:
We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.
Firstly, the above situation can be represented through binomial distribution, i.e.;
where, n = number of samples taken = 150
r = number of success
p = probability of success which in our question is % of bulbs that
are defective, i.e. 10%
<em>Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.</em>
So, Let X = No. of defective bulbs in a box
<u>Mean of X</u>, =
=
= 15
<u>Standard deviation of X</u>, =
=
= 3.7
So, X ~ N(
Now, the z score probability distribution is given by;
Z = ~ N(0,1)
(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X 20) = P(X < 20.5) ---- using continuity correction
P(X < 20.5) = P( <
) = P(Z < 1.49) = 0.9319
(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = =
= 15.
We have been given that there are 125 people and three door prizes.
In the first part we need to figure out how many ways can three door prizes of $50 each be distributed?
Since there are total 125 people and there are three identical door prices, therefore, we need to use combinations for this part.
Hence, the required number of ways are:
In the next part, we need to figure out how many ways can door prizes of $5,000, $500 and $50 be distributed?
Since we have total 125 people and there are three prices of different values, therefore, the required number of ways can be figured out by using permutations.