Let z denote a variable that has a standard normal distribution. Determine the v

Question

Let z denote a variable that has a standard normal distribution. Determine the value z* to satisfy the conditions below. (Round all answers to two decimal places.)

P(z < z*) = 0.0482   z* =

Consider babies born in the "normal" range of 37—43 weeks gestational age. Extensive data support the assumption that for such babies born in the United States, birth weight is normally distributed with mean 3432 g and standard deviation 482 g. (Round your answers to four decimal places.)

What is the probability that the birth weight of a randomly selected baby of this type exceeds 6 lb? (Hint: 1 lb = 453.6 g.)
P(weight > 6 lbs) =

A gasoline tank for a certain car is designed to hold 17.0 gal of gas. Suppose that the variable x = actual capacity of a randomly selected tank has a distribution that is well approximated by a normal curve with mean 17 gal and standard deviation 0.2 gal. (Round all answers to four decimal places.)

What is the probability that a randomly selected tank will hold between 16.4 and 17.3 gal?
P(16.4 x 17.3) =

Suppose that the distribution of net typing rate in words per minute (wpm) for experienced typists can be approximated by a normal curve with mean 60 wpm and standard deviation 25 wpm. (Round all answers to four decimal places.)

What is the probability that a randomly selected typist's net rate is between 10 and 110 wpm?

Suppose that two typists are independently selected. What is the probability that both their typing rates exceed110 wpm?

Suppose that special training is to be made available to the slowest 20% of the typists. What typing speeds would qualify individuals for this training? (Round the answer to one decimal place.)

______________ or less words per minute

Explanation / Answer

Q1.
Mean ( u ) =0
Standard Deviation ( sd )=1
Normal Distribution = Z= X- u / sd ~ N(0,1)                  

P ( Z < x ) = 0.0482
Value of z to the cumulative probability of 0.0482 from normal table is -1.663
P( x-u/s.d < x - 0/1 ) = 0.0482
That is, ( x - 0/1 ) = -1.66
--> x = -1.66 * 1 + 0 = -1.663                  
Q2.
Mean ( u ) =3432
Standard Deviation ( sd )=482
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X > 2721.6) = (2721.6-3432)/482
= -710.4/482 = -1.4739
= P ( Z >-1.474) From Standard Normal Table
= 0.9297

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