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NARA [144]
1 year ago
9

The scores of the students on a standardized test are normally distributed, with a mean of 500 and a standard deviation of 110.

What is the probability that a randomly selected student has a score between 350 and 550? Use the portion of the standard normal table below to help answer the question. z Probability 0.00 0.5000 0.25 0.5987 0.35 0.6368 0.45 0.6736 1.00 0.8413 1.26 0.8961 1.35 0.9115 1.36 0.9131
Mathematics
2 answers:
34kurt1 year ago
7 0
59%................................................................
ololo11 [35]1 year ago
4 0
Μ = 500, population mean
σ = 110, population stadard deviation

The given table is
z  0.00       0.25       0.35      0.45      1.00      1.26       1.35     1.36
P  0.5000  0.5987  0.6368  0.6736  0.8413  0.8961  0.9115  0.9131

Range of random variable is X = [350, 550].

Calculate z-score for x = 350.
z = (350 - 500)/110 = -1.364
From the given tables,
The probability at x = 350 is
1 - 9131 = 0.0869

Calculate the z-score for x = 550.
z = (550 - 500)/110 = 0.454
From the given tables,
The probability at x = 550 is 0.6736

The probability that x =[350,550] is 
0.6736 - 0.0869 = 0.5867

Answer:  0.5867  (or 58.7%)

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Just need to check these answers
amid [387]
Great Job! they are all correct.  :)


Good luck in your next tests.
3 0
2 years ago
Which expression is equivalent to StartFraction 3 x Over x + 1 EndFractiondivided by x + 1?
8_murik_8 [283]

Answer:

Step-by-step explanation:

All those words! Just use math.

\frac{3x/(x+1)}{x+1}  = \frac{3x}{(x+1)^2}

5 0
1 year ago
A downhill skier was able to move 560 meters in 25 seconds. What is the skier’s average speed? (Round your answer to the nearest
salantis [7]
Average speed is the total distance traveled by an object at a certain period of time. To calculate for the average speed we divide the total distance covered by the total time. We do as follows:

Average speed = 560 meters / 25 seconds = 22.4 m/s
6 0
2 years ago
QUESTION THREE (30 MARKS) 3.1 The mass of a standard loaf of white bread is, by law meant to be 700g with a population standard
kiruha [24]

Using the normal distribution and the central limit theorem, it is found that  there is a 0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.

----------------------------------

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

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

Z = \frac{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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

----------------------------------

  • Mean of 700g means that \mu = 700
  • Standard deviation of 21g means that \sigma = 21
  • Sample of 64, thus n = 64
  • <u>For the sampling distribution of the sample mean</u>, the standard deviation is of s = \frac{21}{\sqrt{64}} = \frac{21}{8} = 2.625

The probability of finding a sample mean mass of 695g or below is the p-value of Z when X = 695, thus:

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{695 - 700}{2.625}

Z = -1.905

Z = -1.905 has a p-value of 0.0284.

0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.

A similar problem is given at brainly.com/question/22934264

7 0
1 year ago
Train A and train B stops at Swindon at 10:30 . Train A stops every twelve minutes and train B stops every 14 Mins , when do the
Nata [24]

Answer: at 11:54

Step-by-step explanation:

Let's define the 10:30 as our t = 0 min.

We know that Train A stops every 12 mins, and Train B stops every 14 mins, they will stop at the same time in the least common multiple of 12 and 14.

To find the least common multiple of two numbers, we must do:

LCM(a,b) = a*b/GCD(a,b)

Where GCD(a, b) is the greatest common divisor of a and b.

In this case the only common divisior of 12 and 14 is 2.

So we have:

LCM(12, 14) = 12*14/2 = 84.

Then the both trains will stop 84 minutes after 10:30

one hour has 60 mins, so we can write 84 minutes as:

1 hour and 24 minutes = 1:24

Then they will stop at the same time at 10:30 + 1:24 = 11:54

4 0
1 year ago
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