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melamori03 [73]
3 years ago
8

Calories consumed by members of a track team the day before a race are normally distributed, with a mean of 1,800 calories and a

standard deviation of 100 calories. If a normal curve is sketched using these data, what is the range for 3 standard deviations to the right and to the left of the mean?
0–3,600

1,700–1,900

1,600–2,000

1,500–2,100

_____________________________

The answer is 1,500–2,100

How to find this:

The mean is 1800
The standard deviation is 100

(Subtract 100 from the mean 3 times)
1800 - 100 = 1700
1700 - 100 = 1600
1600 - 100 = 1500

(Add 100 to the mean 3 times)
1800 + 100 = 1900
1900 + 100 = 2000
2000 + 100 = 2100

(The answers are your range)
1500 - 2100

_______________
I just took the test, hope this helps!
Mathematics
1 answer:
irina1246 [14]3 years ago
5 0

Answer: I need help too

Step-by-step explanation:

someone help

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Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let <em>X</em> denote the event that the student has an “on” day, and let <em>Y</em> denote the

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P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})

The events (Y|X) follows a Binomial distribution with probability of success 0.80 and the events (Y|X^{c}) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

P(X)=2\cdot P(X^{c})

Then,

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⇒

2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}

Then,

P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}

Compute the probability that the students passes if request an examination with 3 examiners as follows:

P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})

        =[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}

       =0.715

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})

        =[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}

       =0.734

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

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