Question

1.) the length of human pregnancies has a mean of 266 days and a standard deviation of 16 days. find the probability that a randomly chosen pregnancy lasts between 260 and 270 days. 2.) find the 45th percentile of days for a human pregnancy. 3.) the number of chocolate chips in a bag of chips ahoy cookies has a mean of 1262 with a standard deviation of 118 chips. find the probability a random bag has over 1100 chocolate chips. 4.) True or false. the total area under the normal curve is 5.

Answer 1

Solution :

Given that ,

mean = = 266

standard deviation = = 16

1) P(260 < x < 270) = P[(260 - 266)/ 16) < (x - ) /  < (270 - 266) / 16) ]

= P(-0.38 < z < 0.25)

= P(z < 0.25) - P(z < -0.38)

Using z table,

= 0.5987 - 0.3520

= 0.2467

2) Using standard normal table,

P(Z < z) = 45%

= P(Z < z ) = 0.45

= P(Z < -0.13 ) = 0.45  

z = -0.13

Using z-score formula,

x = z * +

x = -0.13 * 16 + 266

x = 263.92

45 th percentile is =264 days

3) Given that ,

mean = = 1262

standard deviation = = 118

P(x > 1100) = 1 - p( x< 1100)

=1- p P[(x - ) / < (1100 - 1262) / 118]

=1- P(z < -1.37)

Using z table,

= 1 - 0.0853

= 0.9147

4) false. the total area under the normal curve is 0 to 1