Assume that male and female births are equally likely and that the birth of any child does not affect the probability of the gender of any other children find the probability of at most three boys in ten births

Answers

Answer 1

Answer:

The probability of at most three boys in ten births is approximately 0.17139, or about 17.14%.

What is Probability?

It is a branch of mathematics that deals with the occurrence of a random event.

This is a binomial probability problem with n = 10 (number of births) and p = 0.5 (probability of a boy or a girl).

We want to find the probability of at most three boys in ten births, which is equivalent to finding the probability of 0, 1, 2, or 3 boys.

To calculate this probability, we can use the binomial probability formula:

P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= 0.00098 + 0.00977 + 0.04395 + 0.11719

= 0.17139

Therefore, the probability of at most three boys in ten births is approximately 0.17139, or about 17.14%.

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Answer 2

Answer:

Answer:

Step-by-step explanation:

This is a binomial distribution

The probability of at most 3 boys=

P(exactly 0 boys)+P(exactly 1 boy)+P(exactly 2 boys)+P(exactly 3 boys)

.171875

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If f(x)=
$√(x)$
and g(x)=x-9 then what is f(g(13))

Answers

Step-by-step explanation:

f(g(13)) = $√(g(13))$ = $√(13-9)$ = $√(4)$ =2

Assume that females have pulse rates that are normally distributed with a mean of 74.0 beats per minute and a standard deviation of 12.5 beats per minutes. If 16 adult female is randomly selected, find the probability that her pulse rate is less than 80 beats per minute.

Answers

Answer: 0.9726

Step-by-step explanation:

Given : Females have pulse rates that are normally distributed with a mean of 74.0 beats per minute and a standard deviation of 12.5 beats per minutes.

i.e. $\mu=74$ and $\sigma=12.5$

Let x is a random variable to represent the pulse rates.

Formula : $z=(x-\mu)/((\sigma)/(√(n)))$

For n= 16 , the probability that her pulse rate is less than 80 beats per minute will be :-

$P(x<80)=P((x-\mu)/((\sigma)/(√(n)))<(80-74)/((12.5)/(√(16))))\n\n=P(z<(6)/((12.5)/(4)))\n\n=P(z<(24)/(12.5))\n\n=P(z<1.92)=0.9726\ \ [\text{By using z-table.}]$

Hence, the required probability = 0.9726

Final answer:

The probability that a randomly selected female's pulse rate is less than 80 beats per minute, given a mean pulse rate of 74.0 and standard deviation of 12.5 beats per minute, is approximately 0.6844, or 68.44%.

Explanation:

This question pertains to the topic of normal distribution in statistics. We know that the average or mean pulse rate for females is 74.0 beats per minute, with a standard deviation of 12.5 beats per minute. We also know that the pulse rate we want to find the probability for is less than 80 beats per minute.

In these situations, we use the formula for the z-score, which is Z = (X - μ) / σ, where X is the value, we're interested in, μ is the mean, and σ is the standard deviation.

Using this formula, we find Z = (80 - 74) / 12.5 = 0.48. After finding the z-score, we can look at the standard normal distribution table to get the probability. The value for Z = 0.48 on the Z table is approximately 0.6844. Therefore, the probability that a randomly selected female's pulse rate is less than 80 beats per minute is approximately 0.6844, or 68.44%.

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2. From the definition of the derivative (definition 5.3), find the derivative for each of the following functions: (a) f(x) = 6x (b) f(x) = 12x – 2 (c) f(x) = kx² for k a constant