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Scores on a college entrance exam are normally distributed with a mean of 550 and a standard deviation of 100. Find the value that represents the 90th percentile of scores. Answer with a whole number.

Answers

Answer 1

Answer:

Answer:

The value that represents the 90th percentile of scores is 678.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = (X - \mu)/(\sigma)$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 550, \sigma = 100$

Find the value that represents the 90th percentile of scores.

This is the value of X when Z has a pvalue of 0.9. So X when Z = 1.28.

$Z = (X - \mu)/(\sigma)$

$1.28 = (X - 550)/(100)$

$X - 550 = 100*1.28$

$X = 678$

The value that represents the 90th percentile of scores is 678.

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Every day a kindergarten class chooses randomly one of the 50 state flags to hang on the wall, without regard to previous choices, We are interested in the flags that are chosen on Monday, Tuesday and Wednesday of next week.a) Describe a sample space \Omega and a probability measure P to model this experiment.

b) What is the probability that the class hangs Wisconsin's flag on Monday, Michigan's flag on Tuesday, and California's flag on Wednesday.?

c) What is the probability that Wisconsin's flag will be hung at least two of the three days?

Answers

Answer:

a.) P(x = X) = $(1)/(50)$

b.) $(1)/(50) *(1)/(50) *(1)/(50) = (1)/(125000)$

c.) 0.00118

Step-by-step explanation:

The sample space Ω = flags of all 50 states

a.) Any one of the flags is randomly chosen. Therefore we can write the

probability measure as P(x = X) = $(1)/(50)$ , for all the elements of the sample

space, that is for all x ∈ Ω.

b.) the probability that the class hangs Wisconsin's flag on Monday,

Michigan's flag on Tuesday, and California's flag on Wednesday

= $(1)/(50) *(1)/(50) *(1)/(50) = (1)/(125000)$

c.) the probability that Wisconsin's flag will be hung at least two of the three days

= Probability that Wisconsin's flag will be hung on two days + Probability that Wisconsin's flag will be hung on three days

= P(x = 2) + P(x = 3)

= $(\binom{3}{2}* (1)/(50) * (1)/(50)* (49)/(50)) + (\binom{3}{3}* (1)/(50) * (1)/(50)* (1)/(50))\n$

= $(147)/(125000) + (1)/(125000)$

= $(148)/(125000)$

= 0.00118

Final answer:

The sample space for this experiment is all the possible combinations of flags from the 50 U.S. states for the three days. The probability of hanging Wisconsin's flag on Monday, Michigan's on Tuesday, and California's on Wednesday is 1/125,000. The probability of hanging Wisconsin's flag at least two of the three days is 294/125,000.

Explanation:

a) The sample space Ω for this experiment comprises of all possible combinations of flags from the 50 U.S. states for the three days. Hence, the total number of outcomes in the sample space Ω would be 50*50*50 = 125,000. Every outcome in this space is equally likely, so the probability measure P would assign a probability of 1/125,000 to each outcome.

b) As each day's choice is independent of the others and each state's flag is equally likely to be chosen, the probability that Wisconsin's flag is hung on Monday, Michigan's flag is hung on Tuesday, and California's flag is hung on Wednesday would be (1/50) * (1/50) * (1/50) = 1/125,000.

c) To find the probability that Wisconsin's flag will be hung at least two of the three days, we have to add the probabilities for the three situations where Wisconsin's flag is hung exactly twice plus the situation where Wisconsin's flag is hung all three days. The final probability would be [(3 * (1/50)² * (49/50)) + (1/50)³] = 294/125,000.

Learn more about Probability here:

brainly.com/question/22962752

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Can I get the answer for 3 4 5 6

Answers

3) x=190, <BOC=85

4) x=177, <TOU=31

5) x=61, <LOM=110

6) x=55, <DOE=117

Bryce wrote the linear regression equation y=1.123x-6.443 to predict the armspan, y, of a person who is x inches tall. Riley has an armspan of 67.5 inches. Which is the best prediction of Riley’s height? A. 54.36 inches
B. 65.84 inches
C. 66.55 inches
D. 69.36 inches

Answers

Answer:

B. 65.84

Step-by-step explanation:

You are given the armspan and a relation between that and height. Put the given value in the equation and solve for the unknown.

... 67.5 = 1.123x -6.443

... 73.943 = 1.123x

... 73.943/1.123 = x = 65.84

Riley's height is predicted to be 65.84 inches.

The sum of x and -15 is 23

Answers

Answer: x=38

Step-by-step explanation:

x+(-15)=23

x-15=23

+15 +15

x=38

When there is a positive and negative sign the negative sign cancels the positive sign.

Answer:

Step-by-step explanation:

23+15=38 because in order to get 23 from -15 a negative number you need to have 38 to silence the -15.

You can run 10 miles in 80 minutes. How far can you run in 2.5 hours?

Answers

Answer:

18.75 miles

Step-by-step explanation:

1 hour = 60 minutes

2.5 Hours = 150 minutes

10 miles _____> 80 minutes

x_______> 150 minutes

x= ( 150 * 10) / 80

= 1500/80

x= 18.75 miles

easy

Answer:

Step-by-step explanation:

10 miles- 80 minutes

?- 150 minutes

1500+80x

X= 10.75 miles

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Answers

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