Question

A recent survey conducted by a foundation reported that 74​% of teens admitted to texting while driving. A random sample of 42 teens is selected. Use the normal approximation to the binomial distribution to answer parts a through e.

a. Calculate the mean and standard deviation for this distribution.

The mean is _________.

​(Round to four decimal places as​ needed.)

The standard deviation is ___________.

b. What is the probability that more than 36 of the 42 teens admit to texting while​ driving?

The probability is __________..

​(Round to four decimal places as​ needed.)

c. What is the probability that exactly 24 of the 42 teens admit to texting while​ driving?

The probability is _____________.

​(Round to four decimal places as​ needed.)

d. What is the probability that 27​, 28​, or 29 of the 42 teens admit to texting while​ driving?

The probability is ___________.

​(Round to four decimal places as​ needed.)

e. What is the probability that fewer than 32 of the 42

teens admit to texting while​ driving?

The probability is ___________.

​(Round to four decimal places as​ needed.)

Answer 1

Part a)

Using Normal Approximation to Binomial
Mean = n * P = ( 42 * 0.74 ) = 31.08
Variance = n * P * Q = ( 42 * 0.74 * 0.26 ) = 8.0808
Standard deviation = √(variance) = √(8.0808) = 2.8427

Part b)
P ( X > 36 )
Using continuity correction
P ( X > n + 0.5 ) = P ( X > 36 + 0.5 ) = P ( X > 36.5 )

X ~ N ( µ = 31.08 , σ = 2.8427 )
P ( X > 36.5 ) = 1 - P ( X < 36.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 36.5 - 31.08 ) / 2.8427
Z = 1.91
P ( ( X - µ ) / σ ) > ( 36.5 - 31.08 ) / 2.8427 )
P ( Z > 1.91 )
P ( X > 36.5 ) = 1 - P ( Z < 1.91 )
P ( X > 36.5 ) = 1 - 0.9719
P ( X > 36.5 ) = 0.0281

Part c)
P ( X = 24 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 24 - 0.5 < X < 24 + 0.5 ) = P ( 23.5 < X < 24.5 )

X ~ N ( µ = 31.08 , σ = 2.8427 )
P ( 23.5 < X < 24.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 23.5 - 31.08 ) / 2.8427
Z = -2.67
Z = ( 24.5 - 31.08 ) / 2.8427
Z = -2.31
P ( -2.67 < Z < -2.31 )
P ( 23.5 < X < 24.5 ) = P ( Z < -2.31 ) - P ( Z < -2.67 )
P ( 23.5 < X < 24.5 ) = 0.0104 - 0.0038
P ( 23.5 < X < 24.5 ) = 0.0067

Part d)
P ( 27 <= X <= 29 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 27 - 0.5 < X < 29 + 0.5 ) = P ( 26.5 < X < 29.5 )

X ~ N ( µ = 31.08 , σ = 2.8427 )
P ( 26.5 < X < 29.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 26.5 - 31.08 ) / 2.8427
Z = -1.61
Z = ( 29.5 - 31.08 ) / 2.8427
Z = -0.56
P ( -1.61 < Z < -0.56 )
P ( 26.5 < X < 29.5 ) = P ( Z < -0.56 ) - P ( Z < -1.61 )
P ( 26.5 < X < 29.5 ) = 0.2877 - 0.0537
P ( 26.5 < X < 29.5 ) = 0.234

Part e)
P ( X < 32 )
Using continuity correction
P ( X < n - 0.5 ) = P ( X < 32 - 0.5 ) = P ( X < 31.5 )
X ~ N ( µ = 31.08 , σ = 2.8427 )
P ( X < 31.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 31.5 - 31.08 ) / 2.8427
Z = 0.15
P ( ( X - µ ) / σ ) < ( 31.5 - 31.08 ) / 2.8427 )
P ( X < 31.5 ) = P ( Z < 0.15 )
P ( X < 31.5 ) = 0.5596