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3 years ago

The probability that a randomly selected 5-year-old male horned beetle will live to be 6 years old is 0.25624.

Mathematics

2 answers:

dalvyx [7]3 years ago

8 0

Answer:

(a) The probability that two randomly selected 5-year-old male horned beetles will live to be 6 years old is 0.06566.

(b) The probability that six randomly selected 5-year-old male horned beetles will live to be 6 years old is 0.000283.

Step-by-step explanation:

It is given that the probability that a randomly selected 5-year-old male horned beetle will live to be 6 years old is 0.25624.

(a)

We need to find the probability that two randomly selected 5-year-old male horned beetles will live to be 6 years old.

$P=0.25624\times 0.25624=(0.25624)^2=0.0656589376\approx 0.06566$

Therefore the probability that two randomly selected 5-year-old male horned beetles will live to be 6 years old is 0.06566.

(b)

We need to find the probability that six randomly selected 5-year-old male horned beetles will live to be 6 years old.

$P=0.25624\times 0.25624\times 0.25624\times 0.25624\times 0.25624\times 0.25624=(0.25624)^6=0.000283061988948\approx 0.000283$

Therefore the probability that six randomly selected 5-year-old male horned beetles will live to be 6 years old is 0.000283.

Vadim26 [7]3 years ago

7 0

Answer: (a) 0.0656

(b) 0.00028

Step-by-step explanation:

Since we have given that

Probability that a randomly selected 5-year old male horned beetle will live to be 6 years old = 0.25624

If there are

Number of male horned beetles = 2

So, probability that two randomly selected 5-year-old male horned beetles will live to be 6 years old is given by

$(0.25624)^2\\\\=0.0656$

If number of male horned beetles = 6

So, probability that six randomly selected 5-year-old male horned beetles will live to be 6 years old is given by

$(0.25624)^6\\\\=0.00028\\$

Hence, (a) 0.0656

(b) 0.00028

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3 years ago

A certain cold remedy has an 88% rate of success of reducing symptoms within 24 hours. Find the probability that in a random sam

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Answer:

The probability of cured people in who took the remedy is 8/9.

Step-by-step explanation:

Success rate of the cold remedy = 88%

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2 years ago

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3 years ago

The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation

fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 197.5, \sigma = 8.3$

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

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

$Z = \frac{210.9 - 197.5}{8.3}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now $n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14$

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

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

By the Central Limit Theorem

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

$Z = \frac{196 - 197.5}{2.14}$

$Z = -0.7$

$Z = -0.7$ has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

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but we need the Jacobian determinant for the reverse transformation (from $(x,y)$ to $(u,v)$ . To do this, notice that

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3 years ago

The probability that a randomly selected​ 5-year-old male horned beetle will live to be 6 years old is 0.25624.​

The probability that a randomly selected 5-year-old male horned beetle will live to be 6 years old is 0.25624.