For a standardized psychology examination intended for psychology majors, the historical data show that scores have a mean of 505 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?

Answer:[tex]P(\bar X>530)=1-0.808=0.192[/tex] 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: [tex]X \sim N(\mu=505,\sigma=170)[/tex]   And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by: [tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex] On this case  [tex]\bar X \sim N(505,\frac{170}{\sqrt{35}})[/tex] 2) Calculate the probability We want this probability: [tex]P(\bar X>530)=1-P(\bar X<530)[/tex] The best way to solve this problem is using the normal standard distribution and the z score given by: [tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex] If we apply this formula to our probability we got this: [tex]P(\bar X >530)=1-P(Z<\frac{530-505}{\frac{170}{\sqrt{35}}})=1-P(Z<0.87)[/tex] [tex]P(\bar X>530)=1-0.808=0.192[/tex]