Question

The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 7.8 cm

a. Find the probability that an individual distance is greater than 217.50 ????

b. Find the probability that the mean for 25 randomly selected distances is greater than 203.20 cm???

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

Answer 1

Part a)

X ~ N ( µ = 205 , σ = 7.8 )
P ( X > 217.5 ) = 1 - P ( X < 217.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 217.5 - 205 ) / 7.8
Z = 1.6
P ( ( X - µ ) / σ ) > ( 217.5 - 205 ) / 7.8 )
P ( Z > 1.6 )
P ( X > 217.5 ) = 1 - P ( Z < 1.6 )
P ( X > 217.5 ) = 1 - 0.9452
P ( X > 217.5 ) = 0.0548

Part b)

X ~ N ( µ = 205 , σ = 7.8 )
P ( X > 203.2 ) = 1 - P ( X < 203.2 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 203.2 - 205 ) / ( 7.8 / √ ( 25 ) )
Z = -1.15
P ( ( X - µ ) / ( σ / √ (n)) > ( 203.2 - 205 ) / ( 7.8 / √(25) )
P ( Z > -1.15 )
P ( X > 203.2 ) = 1 - P ( Z < -1.15 )
P ( X > 203.2 ) = 1 - 0.1243
P ( X > 203.2 ) = 0.8757

Part c)

Since the population follows normal distribution, hence even though the sample size does not exceed​ 30, we use normal distribution.