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skelet666 [1.2K]
4 years ago
5

A Math professor scores all tests with a maximum of 100 points. A student must have an average of 90 on six tests to receive an

"A" in the course. The student receives 75, 85, 90, 95, and 94 on five tests. Is it possible for the student get an "A" grade?
Mathematics
1 answer:
nignag [31]4 years ago
3 0

Answer:

No, he cannot

Step-by-step explanation:

Lets say the student takes the maximum socre on his last test, so he gets 100. Now, his average on the six texts will be:

(75+85+90+95+94+100)/6 = 89.83

So, if taking 100 is not enough to get to the average of 90, this means he would need a score greater than 100, which is impossible as 100 is the maximum. So, the student CAN'T get an A

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a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles  carry just one person.    

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

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P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

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\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60

We have to evaluate the probability that more than 47 cars carry just one person.

P(x \geq 47)

After continuity correction, we will evaluate

P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)

= 1 - P(z \leq -2.6913)

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P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%

0.996 is the probability that more than half out of 92 vehicles carry just one person.

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