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kotykmax [81]
3 years ago
6

Destiny has 6 library books on her desk. One third of them are about animals. How many of Destiny’s library books are about anim

als? Draw a diagram to show your work.
Mathematics
1 answer:
Marta_Voda [28]3 years ago
3 0
| shade  |               |               |       Each rectangle stands for one third so shade 
|  this     |               |               |         only one rectangle.
|    one   |               |               |         And one rectangle has two books in it (you
|             |               |               |         divide 6 by 3 to get 2). 
|             |               |               |
|             |               |               |
|             |               |               |
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A normally distributed population has mean 57,800 and standard deviation 750. Find the probability that a single randomly select
Stels [109]

Answer:

(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411

(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621

Step-by-step explanation:

We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; \mu = 57,800  and  \sigma = 750.

Let X = randomly selected element of the population

The z probability is given by;

           Z = \frac{X-\mu}{\sigma} ~ N(0,1)  

(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)

P(X <= 58,000) = P( \frac{X-\mu}{\sigma} <= \frac{58000-57800}{750} ) = P(Z <= 0.27) = 0.60642

P(X < 57000) = P( \frac{X-\mu}{\sigma} < \frac{57000-57800}{750} ) = P(Z < -1.07) = 1 - P(Z <= 1.07)

                                                          = 1 - 0.85769 = 0.14231

Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .

(b) Now, we are given sample of size, n = 100

So, Mean of X, X bar = 57,800 same as before

But standard deviation of X, s = \frac{\sigma}{\sqrt{n} } = \frac{750}{\sqrt{100} } = 75

The z probability is given by;

           Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)  

Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)

P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)

P(X bar <= 58,000) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{58000-57800}{\frac{750}{\sqrt{100} } } ) = P(Z <= 2.67) = 0.99621

P(X < 57000) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{57000-57800}{\frac{750}{\sqrt{100} } } ) = P(Z < -10.67) = P(Z > 10.67)

This probability is that much small that it is very close to 0

Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .

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14.04 ft^3

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2.4 ft x 4.5 ft x 1.3 = 14.04 ft^3

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