Question

1. According to the Centers for Disease Control and Prevention, we may assume that the heights of boys aged 8 years in the US have a mean height of 127.5cm with a standard deviation of 5.9cm, and we may assume that the distribution of these heights follow a normal distribution.

a. What percentage of such boys have a height more than 135cm?

b. What percentage of such boys have a height less than 125cm?

c. What percentage of such boys have a height

between 120cm and 130cm?

d. A group of 12 8-year old boys are invited to a birthday party. If this group can be considered as randomly selected, find the probability that the mean height of this group is more than 130cm?

please answer all parts of the wuestion i will thumbs up helpful answers and answer all parts thank you!

Answer 1

Solution :

Given that ,

mean = = 127.5

standard deviation = = 5.9

P(x > 135 ) = 1 - P( x < 135 )

= 1- P[(x - ) / < ( 135 - 127.5) / 5.9]

= 1- P(z < 1.27 )

Using z table,

= 1 - 0.8980

= 0.1020

= 10.20%

Answer = 10.20%

b.

P(x < 125 )

= P[(x - ) / < ( 125 - 127.5) / 5.9 ]

= P(z < -0.42 )

Using z table,

= 0.3372

= 33.72%

Answer = 33.72%

c.

P( 120 < x < 130 )

= P[( 120 - 127.5) / 5.9 ) < (x - ) /  < ( 130 - 127.5) / 5.9) ]

= P( -1.27 < z < 0.42 )

= P(z < 0.42) - P(z < -1.27 )

Using z table,

= 0.6628 - 0.1020

= 0.5608

= 56.08%

Answer = 56.08%

d.

n = 12

= 127.5

= / n = 5.9 / 12 = 1.7032

P( > 130 ) = 1 - P( < 130 )

= 1 - P[( - ) / < ( 130 - 127.5) / 1.7032]

= 1 - P(z < 1.47)

Using z table,    

= 1 - 0.9292

= 0.0708

Probability = 0.0708