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3 years ago

How long would it take you to reach 100 for the number of times you did your activity from this graph?

Mathematics

1 answer:

IceJOKER [234]3 years ago

5 0

Answer:

33 1/3 minutes

Step-by-step explanation:

I note x- number of sentences

y=time

when x=3, y=1

x=6, y= 2

I observe that for every 3 sentences I have one more minute ( y increases by 1) so y= x/3

when x=99 i have y=99/3=33 minutes

for x= 100 I have y=100/3minutes= 33 1/3 minutes

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3 years ago

Ocean fishing for billfish is very popular in the Cozumel region of Mexico. In the Cozumel region about 41% of strikes (while tr

Klio2033 [76]

Answer:

a) 59.10% probability that 12 or fewer fish were caught.

b) 99.74% probability that 5 or more fish were caught.

c) 58.84% probability that between 5 and 12 fish were caught.

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 29, p = 0.41$

$\mu = E(X) = np = 29*0.41 = 11.89$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = 2.6486$

Find the following probabilities.

a) 12 or fewer fish were caught.

Using continuity correction, this is $P(X \leq 12 + 0.5) = P(X \leq 12.5)$ , which is the pvalue of Z when X = 12.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{12.5 - 11.89}{2.6486}$

$Z = 0.23$

$Z = 0.23$ has a pvalue of 0.5910

59.10% probability that 12 or fewer fish were caught.

b) 5 or more fish were caught.

Using continuity correction, this is $P(X \geq 5 - 0.5) = P(X \geq 4.5)$ , which is 1 subtracted by the pvalue of Z when X = 4.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4.5 - 11.89}{2.6486}$

$Z = -2.79$

$Z = -2.79$ has a pvalue of 0.0026

1 - 0.0026 = 0.9974

99.74% probability that 5 or more fish were caught.

c) between 5 and 12 fish were caught.

Using continuity correction, this is $P(5 - 0.5 \leq X \leq 12 + 0.5) = P(4.5 \leq X \leq 12.5)$ , which is the pvalue of Z when X = 12.5 subtracted by the pvalue of Z when X = 4.5. So.