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3 years ago

8

David baked 78 cookies for a bake sale. He set aside 12 cookies to share with his volunteers. The remaining cookies are bagged w

ith 3 cookies in each bad. How many bags does David need?

1 answer:

Sholpan [36]3 years ago

7 0

David would need 22 bags.

(78 - 12) <span>÷ 3 = 22</span>

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The distribution of the number of viewers for the American Idol TV broadcasts follows the normal distribution of a mean of 29 mi

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Answer:

a) $P(30$

And we can find this probability on this way:

$P(0.2$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

$P(0.2$

b) $P(X\geq 23)=P(\frac{X-\mu}{\sigma}\geq \frac{23-\mu}{\sigma})=P(Z\geq \frac{23-29}{5})=P(Z\geq -1.2)$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

$P(Z\geq -1.2)=1-P(Z$

c) $P(X> 40)=P(\frac{X-\mu}{\sigma}> \frac{40-\mu}{\sigma})=P(Z\geq \frac{40-29}{5})=P(Z> 2.2)$

And in order to find the probabilitiy we can use tables for the normal standard distribution, excel or a calculator.

$P(Z>2.2)=1-P(Z$

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

2) Part a

Let X the random variable that represent the number of viewers for the American Idol TV of a population, and for this case we know the distribution for X is given by:

$X \sim N(29,5)$

Where $\mu=29$ and $\sigma=5$ with the values on this case representing millions

We are interested on this probability :

$P(30$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(30$

And we can find this probability on this way:

$P(0.2$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(0.2$

Part b

We are interested on this probability :

$P(X\geq 23)$

$P(X\geq 23)=P(\frac{X-\mu}{\sigma}\geq \frac{23-\mu}{\sigma})=P(Z\geq \frac{23-29}{5})=P(Z\geq -1.2)$

And in order to find the probabilitiy we can use tables for the normal standard distribution, excel or a calculator.

$P(Z\geq -1.2)=1-P(Z$

Part c

We are interested on this probability :

$P(X> 40)$

$P(X> 40)=P(\frac{X-\mu}{\sigma}> \frac{40-\mu}{\sigma})=P(Z\geq \frac{40-29}{5})=P(Z> 2.2)$

And in order to find the probabilitiy we can use tables for the normal standard distribution, excel or a calculator.

$P(Z>2.2)=1-P(Z$

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3 years ago