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LUCKY_DIMON [66]
3 years ago
12

For a standardized psychology examination intended for psychology majors, the historical data show that scores have a mean of 50

5 and a standard deviation of 170. The grading process of this year's exam has just begun. The average score of the 35 exams graded so far is 530.What is the probability that a sample of 35 exams will have a mean score of 530 or more if the exam scores follow the same distribution as in the past?
Mathematics
1 answer:
Volgvan3 years ago
3 0

Answer:

P(\bar X>530)=1-0.808=0.192

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the scores, and for this case we know the distribution for X is given by:

X \sim N(\mu=505,\sigma=170)  

And let \bar X represent the sample mean, the distribution for the sample mean is given by:

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})

On this case  \bar X \sim N(505,\frac{170}{\sqrt{35}})

2) Calculate the probability

We want this probability:

P(\bar X>530)=1-P(\bar X

The best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}

If we apply this formula to our probability we got this:

P(\bar X >530)=1-P(Z

P(\bar X>530)=1-0.808=0.192

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