During boot camp, the drill sergeant measured the weight of the men in his unit. he found the average weight of the men to be 142 pounds and the standard deviation 14 pounds. the data is normally distributed. find the interval in which 68% of the data lies. what is the probability that a man picked at random from the unit will weigh more than 170 pounds? that he will weigh less than 128 pounds?

To solve the question we proceed as follows:
mean=142
standard deviation=14
a] Find the interval in which 68% of the data lies:
P(x<X)=68%=0.68
the z-score associated with this probability is:
P(z<Z)=0.47
but :
z=(x-mu)/sig
thus;
0.47=(x-142)/14
solving for x we get:
x=148.58
thus 68 percent of the data lie below 148.58

b]what is the probability that a man picked at random from the unit will weigh more than 170 pounds?
x=170
thus
P(x>170) will be:
z=(170-142)/14
z=2
Thus
P(x>170)=1-P(z<2)
=1-0.9772
=0.0228

c] that he will weigh less than 128 pounds?
P(x<128)
z=(128-142)/14
z=(-14/14)=-1
Thus
P(z<-1)=0.1587

Answer x=\frac{4(7+y)}{5}x=​5​​4(7+y)​​x=\frac{4(7+y)}{5}x=​5​​4(7+y)​​x=\frac{4(7+y)}{5}x=​5​​4(7+y)​​x=\frac{4(7+y)}{5}x=​5​​4(7+y)​​x=\frac{4(7+y)}{5}x=​5​​4(7+y)​​x=\frac{4(7+y)}{5}x=​5​​4(7+y)​​

step-by-step explanation:

The correct answer is x=-1, usually the answer would be y=something but because there is no slope or y intercept, the answer is x=-1