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Virty [35]
3 years ago
6

If a couple were planning to have three​ children, the sample space summarizing the gender outcomes would​ be: bbb,​ bbg, bgb,​

bgg, gbb,​ gbg, ggb, ggg.
a. construct a similar sample space for the possible weightweight outcomes​ (using o for overweight and u for underweighto for overweight and u for underweight​) of two children.
b. assuming that the outcomes listed in part​ (a) were equally​ likely, find the probability of getting two underweightunderweight children.
c. find the probability of getting exactly one overweightoverweight
Mathematics
1 answer:
MaRussiya [10]3 years ago
5 0
There are two choices for each child: overweight (o) or underweight (u). So if the first child is o the next can be o or u. If the first is u the second can be o or u. This gives four possibilities. Here the first child is the letter noted first and the second is the one listed second:
OO
OU
UO
UU

There are 4 outcomes and if each is equally likely then the probability of each is 1/4. Thus the probability of UU is 1/4

The probability of one underweight and one over weight is 1/2 because in two of the outcomes listed above there is one O and one U (namely OU and UO). Since there are 4 outcomes the probability is 2/4 = 1/2

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The average length of a field goal in the National Football League is 38.4 yards, and the s. d. is 5.4 yards. Suppose a typical
Tasya [4]

Answer:

a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b) 0.25% probability that his average kicks is less than 36 yards

c) 0.11% probability that his average kicks is more than 41 yards

d-a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

d-b) 1.32% probability that his average kicks is less than 36 yards

d-c) 0.80% probability that his average kicks is more than 41 yards

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 problem, we have that:

\mu = 38.4, \sigma = 5.4, n = 40, s = \frac{5.4}{\sqrt{40}} = 0.8538

a. What is the distribution of the sample mean? Why?

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b. What is the probability that his average kicks is less than 36 yards?

This is the pvalue of Z when X = 36. So

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

By the Central Limit Theorem

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

Z = \frac{36 - 38.4}{0.8538}

Z = -2.81

Z = -2.81 has a pvalue of 0.0025

0.25% probability that his average kicks is less than 36 yards

c. What is the probability that his average kicks is more than 41 yards?

This is 1 subtracted by the pvalue of Z when X = 41. So

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

Z = \frac{41 - 38.4}{0.8538}

Z = 3.05

Z = 3.05 has a pvalue of 0.9989

1 - 0.9989 = 0.0011

0.11% probability that his average kicks is more than 41 yards

d. If the sample size is 25 in the above problem, what will be your answer to part (a) , (b)and (c)?

Now n = 25, s = \frac{5.4}{\sqrt{25}} = 1.08

So

a)

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

b)

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

Z = \frac{36 - 38.4}{1.08}

Z = -2.22

Z = -2.22 has a pvalue of 0.0132

1.32% probability that his average kicks is less than 36 yards

c)

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

Z = \frac{41 - 38.4}{1.08}

Z = 2.41

Z = 2.41 has a pvalue of 0.9920

1 - 0.9920 = 0.0080

0.80% probability that his average kicks is more than 41 yards

4 0
3 years ago
What is the least common multiple of 8 and 9?<br><br><br><br> Enter your answer in the box.
Contact [7]

Answer:

72

Step-by-step explanation:

The LCM is when you find the lowest factor that 2 or more numbers share.

For 8 and 9,you would just need to keep multiplying each other until you find the same numbers. It would look something like this.

8,16,24,32,40,48,56,64,72..... etc.

9,18,27,36,45,54,63,72,81..... etc.

You can see how they both have the same factor (72), and since that is the lowest factor (which means I can find a lower factor then 72) then your LCM of 8 and 9 should be 72 :)

4 0
3 years ago
2 6/8 in simplest form
Vesna [10]

Answer:

2 3/4

Step-by-step explanation:

8 0
3 years ago
Find the measure of
Furkat [3]

Answer:

B. 30

Step-by-step explanation:

3x + 2x + x = 180 (angle sum of a triangle is 180)

6x = 180

x = 30

6 0
2 years ago
Read 2 more answers
what is the probability of drawing two blue cards if the first card is not included in the second draw? simplify your answer com
BlackZzzverrR [31]

Answer:

5/14

Step-by-step explanation:

The total number of cards is 8

There are 5 blue cards

P(1st card is blue) = blue/total = 5/8

We take away a blue card

The total number of cards is 7

There are 4 blue cards

P(2nd card is blue) = blue/total = 4/7

The probability of blue blue is P(blue blue ) =

P(1st card is blue)*P(2nd card is blue) =

5/8* 4/7

20/56

We can divide the top and bottom by 4

5/14

3 0
3 years ago
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