Question

The mean score of a college entrance test is 500 ; the standard deviation is 75 . The scores are normally distributed.

a)What percent of the students scored below 320 ?

b) What percentage of the students scored above 575 ?

c) What percentage of the students scored between 400 and 550 ?

Answer 1

I Since it is given the given dota is normally distrubuted so finding the Z-score, 2 x-M xx raw score Mmeon o standard deviation

a) percent of the students scored below 320. Givon, x=320 u=500 T = 75 2= 320-500 75 - 2.40 Now for the percent of the students scored below 320, P(x_320)=P(Z <-2.4) For this open the zotable, split the z-value as = 2.4 and .00 find -2.4 in the column (1st cokon of table) and .00 in the top most saw the corresponding values to these values will be avre percent of students scored below 320; Plx < 320) = P(Z <-24) = 0.0082 6) percent of student scored obove 575. - 75 x=575 u=500 2=575-500 - 1 75 So, we can write P(x>575) = P(Z > 1) = 1 - P(Z < 1)

percent of students scored below Open the z-table, find I in the z codoma (1st cake of table) and at .00 hevel (top most sow), the corresponding value to these two values will be ouse 575, P(Z31) = 0.8413 P(x>575) = P(Z > 1) = 1 -,8413 = 0.1587 c.) percent of students scored between 400 and 550. For this we will find 2, and 22, -1.33 2,- 550-500 0.67 Zi= 400-500 75 75 percent of the students between 400 and 550 P(400 now finding P(Z <-1.33) Split the z-value as, 1.3 ond .03, find -13 in the 2- column and .03 in the top most row, to these 2 value is our P(Z <-1.83); Plz<-1.33) = 0.0918 the contesponding voduce (400 0 0 Add a comment

now finding P(Z <-1.33) Split the z-value as, 1.3 ond .03, find -13 in the 2- column and .03 in the top most row, to these 2 value is our P(Z <-1.83); Plz<-1.33) = 0.0918 the contesponding voduce (400

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