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4 years ago

If a seed is planted, it has a 75% chance of growing into a healthy plant.

Mathematics

1 answer:

Elan Coil [88]4 years ago

6 0

Answer:

9.72405%

Step-by-step explanation:

Binomial Probability

(N choose k) p^k (1-p)^(n-k)

N=7 seeds planted

p= 100% - 70% = 30% = 0.3 <--- we are interested in the plant NOT growing

(1-p) = 70% = 0.7 <--- 70% chance the plant will survive and grow

k=4 <--- we want four of them to fail

The probability is:

(7 choose 4) * (0.3)^4 (0.7)^3 =

7!/(4!3!) (0.3)^4 (0.7)^3 =

(7*6*5/3*2) (0.3)^4 (0.7)^3 =

7*5 (0.3)^4 (0.7)^3 =

35 * 0.0081 * 0.343 = 0.0972405 = 9.72405%

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3 years ago

In a random sample of 25 people, the mean commute time to work was 33.9 minutes and the standard deviation was 7.2 minutes. Ass

Phoenix [80]

Answer:

$33.9-1.32\frac{7.2}{\sqrt{25}}=31.999$

$33.9+1.32\frac{7.2}{\sqrt{25}}=35.801$

So on this case the 80% confidence interval would be given by (31.999;35.801)

And the margin of error is given by:

$ME = t_{\alpha/2}\frac{s}{\sqrt{n}}$

And replacing we got:

$ME = 1.32\frac{7.2}{\sqrt{25}} =1.9008$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=33.9$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=7.2 represent the sample standard deviation

n=25 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=25-1=24$

Since the Confidence is 0.80 or 80%, the value of $\alpha=0.2$ and $\alpha/2 =0.1$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,24)".And we see that $t_{\alpha/2}=1.32$

Now we have everything in order to replace into formula (1):

$33.9-1.32\frac{7.2}{\sqrt{25}}=31.999$

$33.9+1.32\frac{7.2}{\sqrt{25}}=35.801$

So on this case the 80% confidence interval would be given by (31.999;35.801)

And the margin of error is given by:

$ME = t_{\alpha/2}\frac{s}{\sqrt{n}}$

And replacing we got:

$ME = 1.32\frac{7.2}{\sqrt{25}} =1.9008$

7 0

4 years ago