make sure I can see how you did your work and how you arrive

make sure I can see how you did your work and how you arrived at your answers. Question 1: The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with a mean of 266 days and standard deviation of 16 days. a. What percent of pregnancies last less than 240 days? b. What percent of pregnancies last between 240 days and 270 days? c.How long do the longest 20% of pregnancies last? Question 2: A federal report find that a lie detector test given to truthful persons have a probability of 0.2 of suggesting that the person is deceptive. A company asks 12 job applicants to take a lie detect test. Suppose that all 12 applicants answer truthfully. a. What is the probability that exactly one is being deceptive. b.What is the probability that at most one is being deceptive. c.What is the mean and standard deviation of this distribution.

Solution

1)

a) Mean = 266

SD = 16 days

P ( X < 240) = P (Z < 240 - 266 / 16)

= P ( Z < -26 /16 )

= P ( Z < -1.625 )

= 0.0520

= 5.20%

b)

P ( 240 < X < 270)

= P ( 240 - 266 / 16 < Z < 270 - 266 / 16 )

= P ( Z < 0.25) - P ( Z < -1.625)

= 0.5987 - 0.0520

= 0.5367

= 53.67%

c) P(X < x) = 0.80

then x = 279.866

i.e longest 20% of the pregnancy lasts for over 280 days.

2)

a) 1 out of 12 has to be deceptive

Probability of it showing deceptive = 0.2

So, probability of it showing non-deceptive = 0.8

Then, the probability that exactly one is being deceptive = ¹²C₁ (0.2) (0.8)¹¹ = 0.2061

b) Probability of at most one being defective = P (No deceptive) + P (One deceptive)

= ¹²C₀ (0.8)¹² + ¹²C₁ (0.2) (0.8)¹¹

= 0.0687 + 0.2091

= 0.2778

c) For a binomial distribution with n trials

mean = np = 12 * 0.2 = 2.4

variance = np(1-p) = 12 * 0.2 * 0.8 = 1.92

SD = sqrt(1.92) = 1.3856

Hope this helps.