Question

Exercise 1:
The average lifespan of a certain electronic component is 850 days and a standard deviation of 40 days. Knowing that the duration is normally distributed, calculate the probability that this component will last:
a) between 700 and 1000 days;
b) more than 800 days;
c) less than 750 days.

Exercise 2:
A survey reports that 86% of users use Internet Explorer as a browser. You randomly select 200 users and ask them if they use Internet Explorer as a browser. What is the probability that exactly 176 people will answer yes?

Answer 1

1)

a)

X ~ N ( µ = 850 , σ = 40 )
P ( 700 < X < 1000 ) = ?
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 700 - 850 ) / 40
Z = -3.75
Z = ( 1000 - 850 ) / 40
Z = 3.75
P ( -3.75 < Z < 3.75 )
P ( 700 < X < 1000 ) = P ( Z < 3.75 ) - P ( Z < -3.75 )
P ( 700 < X < 1000 ) = 0.9999 - 0.0001
P ( 700 < X < 1000 ) = 0.9998

b)

P ( X > 800 ) = 1 - P ( X < 800 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 800 - 850 ) / 40
Z = -1.25
P ( ( X - µ ) / σ ) > ( 800 - 850 ) / 40 )
P ( Z > -1.25 )
P ( X > 800 ) = 1 - P ( Z < -1.25 )
P ( X > 800 ) = 1 - 0.1056
P ( X > 800 ) = 0.8944

c)

P ( X < 750 ) = ?
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 750 - 850 ) / 40
Z = -2.5
P ( ( X - µ ) / σ ) < ( 750 - 850 ) / 40 )
P ( X < 750 ) = P ( Z < -2.5 )
P ( X < 750 ) = 0.0062

2)

Using Normal Approximation to Binomial
Mean = n * P = ( 200 * 0.86 ) = 172
Variance = n * P * Q = ( 200 * 0.86 * 0.14 ) = 24.08
Standard deviation = √(variance) = √(24.08) = 4.9071

P ( X = 176 ) = ?
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 176 - 0.5 < X < 176 + 0.5 )

= P ( 175.5 < X < 176.5 )

Standardizing the value
Z = ( X - µ ) / σ
Z = ( 175.5 - 172 ) / 4.9071
Z = 0.71
Z = ( 176.5 - 172 ) / 4.9071
Z = 0.92
P ( 0.71 < Z < 0.92 )
P ( 175.5 < X < 176.5 ) = P ( Z < 0.92 ) - P ( Z < 0.71 )
P ( 175.5 < X < 176.5 ) = 0.8212 - 0.7611
P ( 175.5 < X < 176.5 ) = 0.0601