A set of final examination grades in an introductory statist

A set of final examination grades in an introductory statistics course is normally distributed, with a mean of

73 and a standard deviation of 9.

Complete parts (a) through (d).Click here to view page 1 of the cumulative standardized normal distribution table.

a.What is the probability that a student scored below

84 on this exam? The probability that a student scored below 84 is___

b.What is the probability that a student scored between 64 and 94? The probability that a student scored between

64 and 94 is__ .

(Round to four decimal places as needed.)

c. The probability is 25% that a student taking the test scores higher than what grade?The probability is

25% that a student taking the test scores higher than__

(Round to the nearest integer as needed.)

Solution

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    84      
u = mean =    73      
          
s = standard deviation =    9      
          
Thus,          
          
z = (x - u) / s =    1.222222222      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   1.222222222   ) =    0.889188199 [ANSWER]

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b)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    64      
x2 = upper bound =    94      
u = mean =    73      
          
s = standard deviation =    9      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1      
z2 = upper z score = (x2 - u) / s =    2.333333333      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254      
P(z < z2) =    0.990184671      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.831529417   [ANSWER]

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c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.75      
          
Then, using table or technology,          
          
z =    0.67448975      
          
As x = u + z * s,          
          
where          
          
u = mean =    73      
z = the critical z score =    0.67448975      
s = standard deviation =    9      
          
Then          
          
x = critical value =    79.07040775 = 79 [ANSWER]