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shamiya15
20.03.2020 •
Mathematics
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.
a. Find the probability that this box will contain at most 20 defective light bulbs. Show your work or calculator input. (Round your answer to 4 places after the decimal point).
b. How many defective light bulbs are expected to be found in such boxes, on average
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Ответ:
(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%
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.
So, Let X = No. of defective bulbs in a box
Mean of X,
=
=
= 15
Standard deviation of X,
=
=
= 3.7
So, X ~ N(![\mu = 15, \sigma^{2} = 3.7^{2})](/tpl/images/0554/9721/386b1.png)
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.
Ответ:
(C) {8, 20, 30, 35, 42}
Step-by-step explanation:
The given data set is:
14, 30, 35, 8, 29, 28, 31, 42, 20, 36, 32
In order to find the median, first arrange the given data set in ascending order that is:
8, 14, 20, 28, 29, 30, 31, 32, 35, 36, 42
Now, the median of the given data set is:
Also, the lower Quartile is:
8, 14, 20, 28, 29
Thus,![\text{LQ}=20](/tpl/images/0194/1142/78056.png)
And, the upper quartile is:
31, 32, 35, 36, 42
Thus,![\text{UQ}=35](/tpl/images/0194/1142/9b36b.png)
Now, the lower extreme value is 8 and the upper extreme value is 42.
Thus, the list that includes the lower extreme, the lower quartile, the median, the upper quartile, and the upper extreme for the data set given is
{8, 20, 30, 35, 42}.
Therefore, option (C) is correct that is {8, 20, 30, 35, 42}