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erastova [34]
3 years ago
13

Mrs. Farquharson drove through a total of 36 intersections on her way home from work last week. At 4 of every 16 intersections,

Mrs. Farquharson had to stop for a red light before she could drive through. At how many intersections did Mrs. Farquharson have to stop for a red light?
Mathematics
1 answer:
VMariaS [17]3 years ago
7 0
She had to stop at 9
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A study indicates that 62% of students have have a laptop. You randomly sample 8 students. Find the probability that between 4 a
Scrat [10]

Answer:

72.69% probability that between 4 and 6 (including endpoints) have a laptop.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they have a laptop, or they do not. The probability of a student having a laptop is independent from other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

A study indicates that 62% of students have have a laptop.

This means that n = 0.62

You randomly sample 8 students.

This means that n = 8

Find the probability that between 4 and 6 (including endpoints) have a laptop.

P(4 \leq X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 4) = C_{8,4}.(0.62)^{4}.(0.38)^{4} = 0.2157

P(X = 5) = C_{8,5}.(0.62)^{5}.(0.38)^{3} = 0.2815

P(X = 6) = C_{8,6}.(0.62)^{6}.(0.38)^{2} = 0.2297

P(4 \leq X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6) = 0.2157 + 0.2815 + 0.2297 = 0.7269

72.69% probability that between 4 and 6 (including endpoints) have a laptop.

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