Question

Scores of female tests had a mean of 63% (assume normal distribution with a standard deviation of 10%).

A. A girl is randomly selected from all females whose scores are higher than 75%. What is the probability that the girls score is higher than 96%.

B. One thousand people are randomly selected. What is the probability that fewer than 100 of them have a score higher than 75%? Use normal approximation of binomial distribution.

The weight of adult males are normally distributed with a mean of 75kg and a standard deviation of 9kg. What is the probability that 16 randomly selected adult males will have a mean weight of more than 85kg?

Answer 1

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 63
standard Deviation ( sd )= 10
i.
probability that A girl is randomly selected from all females whose scores are higher than 75%
P(X > 75) = (75-63)/10
= 12/10 = 1.2
= P ( Z >1.2) From Standard Normal Table
= 0.1151
=11.51%
ii.
probability that the girls score is higher than 96%
P(X > 96) = (96-63)/10
= 33/10 = 3.3
= P ( Z >3.3) From Standard Normal Table
= 0.0005
=0.05%

b.
NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 1000 * 0.1 = 100
standard deviation ( √npq )= √1000*0.1*0.9 = 9.48683298050514
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
i.
the probability that fewer than 100 of them have a score higher than 75%
P(X > 75) = (75-100)/9.4868
= -25/9.4868 = -2.6352
= P ( Z >-2.6352) From Standard Normal Table
= 0.9958
=99.58%

ii.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 75
standard Deviation ( sd )= 9/ Sqrt ( 16 ) =2.25
sample size (n) = 16

probability that 16 randomly selected adult males will have a mean weight of more than 85kg
P(X > 85) = (85-75)/9/ Sqrt ( 16 )
= 10/2.25= 4.4444
= P ( Z >4.4444) From Standard Normal Table
= 0