Ask question

Ask question

All categories

3 years ago

15

Carrie collected shells from the seashore in a box he takes out a handful of shells from the box is this a random sample of shel

ls in a box

1 answer:

Leya [2.2K]3 years ago

4 0

This would be random. This is because Carrie has no clue or has a limited knowledge of what type of shells he would grab.

You might be interested in

Flip a coin 10 times and record the observed number of heads and tails. For example, with 10 flips one might get 6 heads and 4 t

zepelin [54]

Answer:

The greater the sample size the better is the estimation. A large sample leads to a more accurate result.

Step-by-step explanation:

Consider the table representing the number of heads and tails for all the number of tosses:

Number of tosses n (HEADS) n (TAILS) Ratio

10 3 7 3 : 7

30 14 16 7 : 8

100 60 40 3 : 2

Compute probability of heads for the tosses as follows:

n = 10 tosses

$P(\text{HEADS})=\frac{3}{10}=0.30$

The probability of heads in case of 10 tosses of a coin is -0.20 away from 50/50.

n = 30 tosses

$P(\text{HEADS})=\frac{14}{30}=0.467$

The probability of heads in case of 30 tosses of a coin is -0.033 away from 50/50.

n = 100 tosses

$P(\text{HEADS})=\frac{60}{100}=0.60$

The probability of heads in case of 100 tosses of a coin is 0.10 away from 50/50.

As it can be seen from the above explanation, that as the sample size is increasing the distance between the expected and observed proportion is decreasing.

This happens because, the greater the sample size the better is the estimation. A large sample leads to a more accurate result.

5 0

2 years ago

Svetllana [295]

Answer:

Answer A = 5/2

Answer B = 4

Answer C = -3

Answer D = 3

Answer E = 12/5

6 0

3 years ago

Identify the coefficients and constants in the expression. 2x – y + 5x

garri49 [273]

I believe the answer is 7x-y

7 0

2 years ago

What is greater 8ft or 3yr

timama [110]

3yr is greater than 8ft

3 0

3 years ago

Read 2 more answers

According to a Yale program on climate change communication survey, 71% of Americans think global warming is happening.† (a) For

SpyIntel [72]

Answer:

a) 0.2741 = 27.41% probability that at least 13 believe global warming is occurring

b) 0.7611 = 76.11% probability that at least 110 believe global warming is occurring

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

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:

$p = 0.71$

(a) For a sample of 16 Americans, what is the probability that at least 13 believe global warming is occurring?

Here $n = 16$ , we want $P(X \geq 13)$ . So

$P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 13) = C_{16,13}.(0.71)^{13}.(0.29)^{3} = 0.1591$

$P(X = 14) = C_{16,14}.(0.71)^{14}.(0.29)^{2} = 0.0835$

$P(X = 15) = C_{16,15}.(0.71)^{15}.(0.29)^{1} = 0.0273$

$P(X = 16) = C_{16,16}.(0.71)^{16}.(0.29)^{0} = 0.0042$

$P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) = 0.1591 + 0.0835 + 0.0273 + 0.0042 = 0.2741$

0.2741 = 27.41% probability that at least 13 believe global warming is occurring

(b) For a sample of 160 Americans, what is the probability that at least 110 believe global warming is occurring?

Now $n = 160$ . So

$\mu = E(X) = np = 160*0.71 = 113.6$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{160*0.71*0.29} = 5.74$

Using continuity correction, this is $P(X \geq 110 - 0.5) = P(X \geq 109.5)$ , which is 1 subtracted by the pvalue of Z when X = 109.5. So

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

$Z = \frac{109.5 - 113.6}{5.74}$

$Z = -0.71$

$Z = -0.71$ has a pvalue of 0.2389

1 - 0.2389 = 0.7611

0.7611 = 76.11% probability that at least 110 believe global warming is occurring

3 0

2 years ago