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stich3 [128]
1 year ago
12

Scores on an exam follow an approximately normal distribution with a mean of 76. 4 and a standard deviation of 6. 1 points. What

is the minimum score you would need to be in the top 2%?.
Mathematics
1 answer:
crimeas [40]1 year ago
5 0

If scores on an exam follow an approximately normal distribution with a mean of 76.4 and a standard deviation of 6.1 points, then the minimum score you would need to be in the top 2% is equal to 88.929.

A problem of this type in mathematics can be characterized as a normal distribution problem. We can use the z-score to solve it by using the formula;

Z = x - μ / σ

In this formula the standard score is represented by Z, the observed value is represented by x, the mean is represented by μ, and the standard deviation is represented by σ.

The p-value can be used to determine the z-score with the help of a standard table.

As we have to find the minimum score to be in the top 2%, p-value = 0.02

The z-score that is found to correspond with this p-value of 0.02 in the standard table is 2.054

Therefore,

2.054 = x - 76.4 ÷ 6.1

2.054 × 6.1 = x - 76.4

12.529 = x - 76.4

12.529 + 76.4 = x

x = 88.929

Hence 88.929 is calculated to be the lowest score required to be in the top 2%.

To learn more about normal distribution, click here:

brainly.com/question/4079902

#SPJ4

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After driving to a riverfront parking lot, Bob plans to run south along the river, turn around, and return to the parking lot, r
mars1129 [50]

Answer:A) 1.5

Step-by-step explanation: Bob runs at the rate of 8mins per mile

In 60mins his rate would be=60/8=7.5

Let a be the distance he further runs south

2s+3.25

Total distance covered in 50mins=Time=distance/speed=

50/60

50/60=2s +3.25/7.5

Cross multiply

60(2s+3.25)=50×7.5

120s+195=375

120s=375-195

S=180/120

S=1.5

5 0
3 years ago
Researchers recorded the speed of ants on trails in their natural environments. The ants studied, Leptogenys processionalis, all
posledela

This question is Incomplete

Complete Question

Researchers recorded the speed of ants on trails in their natural environments. The ants studied, Leptogenys processionalis, all have the same body size in their adult phase, which made it easy to measure speeds in units of body lengths per second (bl/s). The researchers found that, when traffic is light and not congested, ant speeds vary roughly Normally, with mean 6.20 bl/s and standard deviation 1.58 bl/s. (a) What is the probability that an ant's speed in light traffic is faster than 5 bl/s? You may find Table B useful. (Enter your answer rounded to four decimal places.)

Answer:

0.7762

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

Population mean = 6.20 bl/s

Standard deviation = 1.58 bl/s.

x = 5 bl/s

z = 5 - 6.20/1.58

z = -0.75949

The probability that an ant's speed in light traffic is faster than 5 bl/s is P( x > 5)

Probability value from Z-Table:

P(x<5) = 0.22378

P(x>5) = 1 - P(x<5)

= 1 - 22378

= 0.77622

Approximately to 4 decimal places = 0.7762

The probability that an ant's speed in light traffic is faster than 5 bl/s is 0.7762

5 0
3 years ago
At an annual salary of $55,000 per year, what is the amount of semimonthly paychecks?
Amiraneli [1.4K]
Semi monthly means 2 times per month
assume that pay is at beginning and middle of month (doesn't matter, but  makes it easier for me)

so 12 months
2 times per month
12 times 2=24
divide 55000 by number of months
55,000/24=2291.67

so the amount per paycheck, assuming equally distributed amounts throughout the paychecks, is $2291.67
4 0
3 years ago
Helppppppp asappppp
vaieri [72.5K]

Answer:

Your answer is 87.487699

4 0
3 years ago
What are ALL of the factors of 7? Mark ALL that Apply. 7,1,5,3,4,2
lesya692 [45]

Answer:

Just 1 and 7 I believe since it is a prime number :)

3 0
3 years ago
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