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2 years ago

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Some psychologists believe that a "genius" should be defined as anyone having an IQ over 140. If IQ scores are normally distribu

ted with a mean of 100 and a standard deviation of 13, and if the population of the world is 6,575,000,000, how many geniuses are there in the world today? A. 115,000 B. 24,985,000 C. 536,000,000 D. 123,000 E. 6,575,000

1 answer:

Zinaida [17]2 years ago

5 0

I'm positive it's B. 24,985,000

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On average, the number of customers who had items to return for refunds or exchanges at a certain retail store's service desk is

spayn [35]

Answer:

The probability that the service desk will have at least 100 customers with returns or exchanges on a randomly selected day is P=0.78.

Step-by-step explanation:

With the weekly average we can estimate the daily average for customers, assuming 7 days a week:

$M=756/7=108$

We can model this situation with a Poisson distribution, with parameter λ=108. But because the number of events is large, we use the normal aproximation:

$P(\lambda)\approx N(\lambda,\lambda)$

Then we can calculate the z value for x=100:

$z=\frac{x-\mu}{\sigma}=\frac{100-108}{\sqrt{108}}=\frac{-8}{10.4} =-0.77$

Now we calculate the probability of x>100 as:

$P(x>100)=P(z>-0.77)=0.78$

The probability that the service desk will have at least 100 customers with returns or exchanges on a randomly selected day is P=0.78.

5 0

2 years ago

The sum of two polynomials is 8d5 – 3c3d2 + 5c2d3 – 4cd4 + 9. If one addend is 2d5 – c3d2 + 8cd4 + 1, what is the other addend?

Anit [1.1K]

Answer: The correct option is A.

Step-by-step explanation: We are given a polynomial which is a sum of other 2 polynomials.

We are given the resultant polynomial which is : $8d^5-3c^3d^2+5c^2d^3-4cd^4+9$

One of the polynomial which are added up is : $2d^5-c^3d^2+8cd^4+1$

Let the other polynomial be 'x'

According to the question:

$8d^5-3c^3d^2+5c^2d^3-4cd^4+9=x+(2d^5-c^3d^2+8cd^4+1)$

$x=8d^5-3c^3d^2+5c^2d^3-4cd^4+9-(2d^5-c^3d^2+8cd^4+1)$

Solving the like terms in above equation we get:

$x=(8d^5-2d^5)+(-3c^3d^2+c^3d^2)+(5c^2d^3)+(-4cd^4-8cd^4)+(9-1)$

$x=6d^5-2c^3d^2+5c^2d^3-12cd^4+8$

Hence, the correct option is A.

5 0

2 years ago

Read 2 more answers

A manufacturing company produces valves in various sizes and shapes. One particular valve plate is supposed to have a tensile st

maxonik [38]

Answer:

$z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413$

$p_v =2*P(z>1.413)=0.158$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

Step-by-step explanation:

Data given and notation

$\bar X=5.0611$ represent the sample mean

$\sigma=0.2803$ represent the population standard deviation for the sample

$n=42$ sample size

$\mu_o =5$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is different from 5, the system of hypothesis would be:

Null hypothesis: $\mu = 5$

Alternative hypothesis: $\mu \neq 5$

If we analyze the size for the sample is > 30 but and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413$

P-value

Since is a two sided test the p value would be:

$p_v =2*P(z>1.413)=0.158$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

8 0

2 years ago

If a is an arbitrary nonzero constant, what happens to a/b as b approaches 0

Stolb23 [73]

It depends on how b approaches 0

If b is positive and gets closer to zero, then we say b is approaching 0 from the right, or from the positive side. Let's say a = 1. The equation a/b turns into 1/b. Looking at a table of values, 1/b will steadily increase without bound as positive b values get closer to 0.

On the other side, if b is negative and gets closer to zero, then 1/b will be negative and those negative values will decrease without bound. So 1/b approaches negative infinity if we approach 0 on the left (or negative) side.

The graph of y = 1/x shows this. See the diagram below. Note the vertical asymptote at x = 0. The portion to the right of it has the curve go upward to positive infinity as x approaches 0. The curve to the left goes down to negative infinity as x approaches 0.

7 0

2 years ago

Suppose that the weight of navel oranges is normally distributed with a mean µµ = 8 ounces, and a standard deviation σσ = 1.5 ou

monitta

Answer:

Hello some parts of your question is missing below is the missing part

c. If you randomly select a navel orange, what is the probability that it weighs between6.2 and 7 ounces

Answer: A) 0.0099

B) 0.6796

C) 0.13956

Step-by-step explanation:

weight of Navel oranges evenly distributed

mean ( u ) = 8 ounces

std ( б )= 1.5

navel oranges = X

A ) percentage of oranges weighing more than 11.5 ounces

P( x > 11.5 ) = $P ( \frac{x - u}{ std} > \frac{11.5-8}{1.5} )$

= P ( Z > 2.33 ) = 0.0099

= 0.9%

B) percentage of oranges weighing less than 8.7 ounces

P( x < 8.7 ) = $P ( \frac{x - u}{ std} > \frac{8.7-8}{1.5} )$

= P ( Z < 0.4667 ) = 0.6796

= 67.96%

C ) probability of orange selected weighing between 6.2 and 7 ounces?

P ( 6.2 < X < 7 ) = $P (\frac{6.2-8}{1.5} < \frac{x - u}{ std} < \frac{7-8}{1.5} )$

= P ( -1.2 < Z < -0.66 )

= Ф ( -0.66 ) - Ф(-1.2) = 0.13956

7 0

2 years ago