Columbia manufactures bowling balls with a mean weight of 14.2 pounds and a standard deviation of 1.6 pounds. A bowling ball is too heavy to use and is discarded if it weighs over 16 pounds. Assume that the weights of bowling balls manufactured by Columbia are normally distributed. Round answers to four decimal places. a) What is the probability that a randomly selected bowling ball is discarded due to being too heavy to use? | 1.125 x b) The lightest 6% of the bowling balls made are discarded due to the possibility of defects. A bowling ball is discarded for being too light if it weighs under what specific weight? 12.6452 @ pounds c) What is the probability that a randomly selected bowling ball will be discarded for being either too heavy or too light? 1.185 x

note these answer may differ from the ones you may get by doing  it by software

a)

=0.1292

b)

=11.7200

c)

=0.1892

software answers

a) 0.1303

b) 11.7124

c) 0.1903

Step-by-step explanation

here we to use the concept of  z scores to solve  this question we  know given the mean and standard deviation of a random variable x  then  we can find its z score and the read the probability associated with the z score from the normal bales 

 if we let m denote mean and s  denote standard deviation then iven any raw score x then z=(x-m)/s

from our case above we are given m=14.2 and  s=1.6

a)

we are told if x>16 the we will discard then 

P(x>16)

=P(z>(16-14.2)/1.6)

=P(z>1.13)

=1-P(z<=1.13)

=1-0.8708

=0.1292

b)

here we need to find a z score associated with probability of 0.06 i.e Z_0.06=-1.55

using the formula z=(x-m)/s

thus x=z*s +m

=-1.55*1.6 +14.2

=11.72

c)

for this to occur we need P() or P(x>16)

P(x<11.72)=0.06 from part b and P(x>16)=0.1292 from part a

to get the total probability sum the two

i.e. 

0.06+0.1292

=0.1892