Question

Suppose 31 pregnant women are sampled who smoke an average of 22 cigarettes per day with a variance of 144.00.

a) What is the probability that the pregnant women will smoke an average of 20 cigarettes or more? probability =

b) What is the probability that the pregnant women will smoke an average of 21 cigarettes or less? probability =

c) What is the probability that the pregnant women will smoke an average of 18 to 24 cigarettes? probability =

d) What is the probability that the pregnant women will smoke an average of 23 to 26 cigarettes? probability =

Note: Do NOT input probability responses as percentages; e.g., do NOT input 0.9194 as 91.94.

Answer 1

Solution :

Given that ,

mean = = 22

variance = ² = 144.00

standard deviation = = ² = 144.00 = 12

n = 31

=   = 22

= / n = 12 / 31 = 2.16

a) P( 20) = 1 - P( 20)

= 1 - P[( - ) / (20 - 22) / 2.16]

= 1 - P(z -0.93)

Using z table,    

= 1 - 0.1762

= 0.8238

b) P( 21) = P(( - ) / (21 - 22) /2.16 )

= P(z -0.46)

Using z table

= 0.3228

c) P(18 < < 24)  

= P[(18 - 22) / 2.16 < ( - ) / < (24 - 22) / 2.16 )]

= P( -1.85 < Z < 0.93)

= P(Z < 0.93) - P(Z < -1.85)

Using z table,  

= 0.8238 - 0.0322

= 0.7916

d) P(23 < < 26)  

= P[(23 - 22) / 2.16 < ( - ) / < (26 - 22) / 2.16 )]

= P( 0.46 < Z < 1.85)

= P(Z < 1.85) - P(Z < 0.46)

Using z table,  

= 0.9678 - 0.6772  

= 0.2906