A teacher wishes to \"curve\" a test whose grades were normally distributed with

Question

A teacher wishes to "curve" a test whose grades were normally distributed with a mean of 60 and standard deviation of 14. The top 15% of the class will get an A, the next 25% of the class will get a B, the next 30% of the class will get a C, the next 25% of the class will get a D and the bottom 5% of the class will get an F. Find the cutoff for each of these grades. (Round your answers to two decimal places.) (a) The A cutoff is a grade of (b) The B cutoff is a grade of (c) The C cutoff is a grade of (d) The D cutoff is a grade of TKK Products manufactures 50-, 60-, 75-r and 100-watt electric light bulbs. Laboratory tests show that the lives of these light bulbs are normally distributed with a mean of 550 hr and standard deviation of 150 hr. What is the probability that a TKK light bulb selected at random will burn for the following times (Round your answers to four decimal places.) (a) more than 750 hr (b) less than 450 hr (c) between 550 and 750 hr (d) between 450 and 600 hr

Explanation / Answer

4.              
Mean ( u ) =550
Standard Deviation ( sd )=150
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.                  
P(X > 750) = (750-550)/150
= 200/150 = 1.3333
= P ( Z >1.333) From Standard Normal Table
= 0.0912                  
b.
P(X < 450) = (450-550)/150
= -100/150= -0.6667
= P ( Z <-0.6667) From Standard Normal Table
= 0.2525                  
c.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 550) = (550-550)/150
= 0/150 = 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(X < 750) = (750-550)/150
= 200/150 = 1.3333
= P ( Z <1.3333) From Standard Normal Table
= 0.90879
P(550 < X < 750) = 0.90879-0.5 = 0.4088                  
d.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 450) = (450-550)/150
= -100/150 = -0.6667
= P ( Z <-0.6667) From Standard Normal Table
= 0.25249
P(X < 600) = (600-550)/150
= 50/150 = 0.3333
= P ( Z <0.3333) From Standard Normal Table
= 0.63056
P(450 < X < 600) = 0.63056-0.25249 = 0.3781                  
          

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