For each sunflower, there are only two possible outcomes. Either it germinates, or it does not. The probability of a sunflower germinating is independent of other sunflowers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

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

The standard deviation of the binomial distribution is:

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

Seventy-six percent of sunflower seeds will germinate into a flower

This means that $p = 0.76$

Samples of 800:

This means that $n = 800$

The standard deviation for the number of sunflower seeds that will germinate in such samples of size 800 is:

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{800*0.76*0.24} = 12.08$

5 0

3 years ago

I really need this answered fast.....The coordinates of the vertices of △JKL are J(−5, −1) , K(0, 1) , and L(2, −5) . Which stat

dimulka [17.4K]

<span>JKL is a right triangle because JK ¯ ¯ ¯ ¯ ¯ is perpendicular to KL</span>

5 0

3 years ago

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Recall that for use of a normal distribution as an approximation to the binomial distribution, the conditions np greater than or

vagabundo [1.1K]

Answer:

The minimum sample size needed for use of the normal approximation is 50.

Step-by-step explanation:

Suitability of the normal distribution:

In a binomial distribution with parameters n and p, the normal approximation is suitable is:

np >= 5

n(1-p) >= 5

In this question, we have that:

p = 0.9

Since p > 0.5, it means that np > n(1-p). So we have that:

$n(1-p) \geq 5$

$n(1 - 0.9) \geq 5$

$0.1n \geq 5$

$n \geq \frac{5}{0.1}$

$n \geq 50$

The minimum sample size needed for use of the normal approximation is 50.

6 0

3 years ago