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olganol [36]
3 years ago
5

A consensus forecast is the average of a large number of individual analysts' forecasts. Suppose the individual forecasts for a

particular interest rate are normally distributed with a mean of 5.0 percent and a standard deviation of 1.2 percent. A single analyst is randomly selected. Find the probability that his/her forecast is(a) At least 3.5 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)(b) At most 6 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)(c) Between 3.5 percent and 6 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)
Mathematics
1 answer:
zaharov [31]3 years ago
3 0

Answer:

a) P(X>3.5)=P(\frac{X-\mu}{\sigma}>\frac{3.5-\mu}{\sigma})=P(Z>\frac{3.5-5}{1.2})=P(z>-1.25)

And we can find this probability using the complement rule:

P(z>-1.25)=1-P(z

b) P(X

And we can find this probability uing the normal standard table:

P(z

c) P(3.5

And we can find this probability with this difference:

P(-1.25

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

P(-1.25

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".  

Part a

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(5,1.2)  

Where \mu=5 and \sigma=1.2

We are interested on this probability

P(X>3.5)

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(X>3.5)=P(\frac{X-\mu}{\sigma}>\frac{3.5-\mu}{\sigma})=P(Z>\frac{3.5-5}{1.2})=P(z>-1.25)

And we can find this probability using the complement rule:

P(z>-1.25)=1-P(z

Part b

We are interested on this probability

P(X

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(X

And we can find this probability uing the normal standard table:

P(z

Part c

P(3.5

And we can find this probability with this difference:

P(-1.25

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

P(-1.25

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Irina-Kira [14]

Answer:

a. H0 : p ≤ 0.11 Ha : p >0.11 ( one tailed test )

d. z= 1.3322

Step-by-step explanation:

We formulate our hypothesis as

a. H0 : p ≤ 0.11 Ha : p >0.11 ( one tailed test )

According to the given conditions

p`= 31/225= 0.1378

np`= 225 > 5

n q` = n (1-p`) = 225 ( 1- 31/225)= 193.995> 5

p = 0.4 x= 31 and n 225

c. Using the test statistic

z=  p`- p / √pq/n

d. Putting the values

z= 0.1378- 0.11/ √0.11*0.89/225

z= 0.1378- 0.11/ √0.0979/225

z= 0.1378- 0.11/ 0.02085

z= 1.3322

at 5% significance level the z- value is ± 1.645 for one tailed test

The calculated value falls in the critical region so we reject our null hypothesis H0 : p ≤ 0.11 and accept  Ha : p >0.11 and  conclude that the data indicates that the 11% of the world's population is left-handed.

The rejection region is attached.

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3 years ago
Please help
Ann [662]

Answer:

1/60 probabiliity

Step-by-step explanation:

You have two independent events that you want to put together.

Let Pr. mean "probability"

Pr(5 from 10 cards and 2 on a dice ) = Pr(5 from 10 cards) * Pr( 2 on a dice)

Pr(5 from 10 cards and 2 on a dice ) =(1/10) * (1/6)

= 1/60

Pr(5 from 10 cards and 2 on a dice ) =0.0167

or 1.67% probability

5 0
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