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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 distribute

denis23 [38]

Answer:

61,239,550

Step-by-step explanation:

We let the random variable X denote the IQ scores. This would imply that X is normal with a mean of 100 and standard deviation of 17. We proceed to determine the probability that an individual chosen at random from the population would be a genius, that is;

Pr( X>140)

The next step is to evaluate the z-score associated with the IQ score of 140 by standardizing the random variable X;

$Pr(X>140)=Pr(Z>\frac{140-100}{17})=Pr(Z>2.3529)$

The area to the right of 2.3529 will be the required probability. This area from the standard normal tables is 0.009314

From a population of 6,575,000,000 the number of geniuses would be;

6,575,000,000*0.009314 = 61,239,550

6 0

3 years ago

Read 2 more answers

PLEASE ANSWER 100POINTS+BRAINLEST PLEASE HELP NO DUMMY ANSWERS

makkiz [27]

The answer is on google no dumb stuff dummy

7 0

3 years ago

Read 2 more answers

Jenny bought a pair of pants for $19.99 and two shirts for $7.50. If she gave the cashier 40 dollars how much change did she rec

VMariaS [17]

Answer:

Jenny will receive $12.51 in change.

Step-by-step explanation:

We need to find the total cost first.

19.99 + 7.50 = 27.49

We then subtract that from the 40 dollars she gave to get the amount of change she receives.

40.00 - 27.49 = 12.51.

She will receive $12.51 in change.

hope this helped!

4 0

3 years ago

All bags entering a research facility are screened. The screening process is not perfect so that 77% of the bags that contain fo

garik1379 [7]

Answer:

(1) 0.5933

(2) 0.8955

Step-by-step explanation:

We are given that all bags entering a research facility are screened.

Let Probability that bags entering the building contains forbidden material,

P(F) = 0.69

Probability that bags entering the building does not contains forbidden material, P(NF) = 1 - 0.69 = 0.31

Let event A = alarm gets triggered

Probability that alarm gets trigger given the bags contain forbidden material, P(A/F) = 0.77

Probability that alarm gets trigger given the bags does not contain forbidden material, P(A/NF) = 0.20

(1) Probability that a bag triggers the alarm, P(A) ;

P(A) = P(F) * P(A/F) + P(NF) * P(A/NF)

= (0.69 * 0.77) + (0.31 * 0.20) = 0.5313 + 0.062

= 0.5933

Therefore, probability that a bag triggers the alarm is 0.5933 .

(2) Probability that a bag that triggers the alarm will actually contain forbidden material is given by P(F/A) ;

Using Bayes' Theorem;

P(F/A) = $\frac{P(F) * P(A/F)}{P(F) * P(A/F) + P(NF) *P(A/NF)}$ = $\frac{0.69*0.77}{0.69*0.77+0.31*0.20}$ = $\frac{0.5313}{0.5933}$

= 0.8955

6 0

3 years ago

Many newspapers carmy a certain purzile in which the reader must unsoramble leers to form words how many warys cam the letters o

Kruka [31]

The correct unscrambling of whyrot is worthy.

<h3>What is unscrambling?</h3>

A word unscrambler is essentially a gadget that you feed all the letters into and it rearranges them to reveal all potential word combinations.
Some might be concerned that this is a cheating method. However, if everyone playing the game has the option of using a word unscrambler, then the playing field is unquestionably even.
A player may opt not to use the word generator and instead come up with their own list of terms. Having said that, they might want to utilize it afterward to assess their performance and review the entire list of terms they could have used.

To learn more about unscrambling with the given link

brainly.com/question/21327199

#SPJ4

3 0

1 year ago