Question

The certain paper suggested that a normal distribution with mean 3,500 grams and standard deviation 590 grams is a reasonable model for birth weights of babies born in Canada. (Use a table or technology.) (a) One common medical definition of a large baby is any baby that weighs more than 4,000 grams at birth. What is the probability that a randomly selected Canadian baby is a large baby? (Round your answer to four decimal places.) (b) What is the probability that a randomly selected Canadian baby weighs either less than 2,000 grams or more than 4,000 grams at birth? (Round your answer to four decimal places.) (c) What birth weights describe the 25% of Canadian babies with the greatest birth weights? (Round your answer to the nearest integer.) Any baby with a weight of grams or more is in the greatest 25% of birth weights.

Answer 1

1)Solution:
Given that,
μ =3500 grams,σ=590 grams
A) P(X>4000)= 1-P(X<=4000)
=1-p{[(x- μ)/σ]<=[(4000- 3500)/590]}
=1-P(z<=0.85)
=1-0.8023 ( from Standard Normanl table)
P(X>4000)= 0.1977
B) P[(X<2000) or (X>4000)]=P(X<2000) +P(X>4000)
= p{[(x- μ)/σ]<[(2000 - 3500)/590]} + 1-P(X<=4000)
=P(z< -2.54)+1-p{[(x- μ)/σ]<=[(4000- 3500)/590]}

=0.0055+[1-p(z<=0.85)]

=0.0055+(1-0.8023)

=0.0055+0.1977
=0.2032
P(X<2000) or P(X>4000)=0.2032
C)By using standard normal distribution ,we have to find the corresponding z-values as
P( Z> z)= 25%
1-p( Z<= z)= 0.25
P( Z<=0.67)= 0.75
From the standard normal table.
z= 0.67
Now using the z-score formula, we can find the value of x as follows,
z=(x- μ)/σ
z×σ=( x- μ)
x= μ+( z×σ)
x=3500+(0.67×590)
=3500+395.3
x=3895.30
Any baby with a weight of 3895.30 grams or more is the greatest 25% of the birth weight.
2) Solution:
Given that,
μ =0.498 inches,σ=0.002 inches
A machine that produces a ball bearings is acceptable if it's diameter is within 0.004 inches of mean diameter 0.500 inches as a target value.
P(the bearing produced is not acceptable)

=1-P(the bearing produced is acceptable)
=1-p(0.500-0.004<X<0.500-0.004)
=1-p(0.496<X<0.504)
=1- p{[(0.496-0.498)/0.002]<[(x- μ)/σ]<[(0.504 - 0.498)/0.002]}
=1-P(-1<Z< 3)
=1-[ p(Z< 3) - p(Z< -1)]
=1-[0.9987 - 0.1587] ( from. Standard Normanl table)
=1-0.8400
=0.16
P(the bearing produced is not acceptable)=0.16
In percentage,
P(the bearing produced is not acceptable)=16%
3) solution:
By using the standard normal table, we can find the following probabilities as,
a) p( Z< -1.26)= 0.1038
The area to the left of -1. 26 is 0.1038
b) p( Z>1.26)=1- p( Z<=1.26)
=1-0.8962
=0.1038
The area to the right of 1. 26 is 0.1038
C) P(-1<Z< 2)=p(Z< 2) - p(Z< -1)
= [0.9772 - 0.1587] ( from. Standard Normanl table)
=0.8185
The area in between -1 to 2 is 0.8185
D) p( Z> 0)=1- p( Z<=0)
=1-0.5
=0.5
The area to the right of 0 is 0.5
E) p( Z> -6)=1- p( Z<= -6)
=1-0
=1
The area to the right of -6 is 1
F) P(-1.4<Z< 2.6)=p(Z< 2.6) - p(Z< -1.4)
= 0.9953 - 0.0808 ( from. Standard Normanl table)
=0.9145
The area in between -1.4 to 2.6 is 0.9145
G) p( Z< 0.24)= 0.5948
The area to the left of 0.24 is 0.5948