Statistics questions

Question

The mean incubation time for a type of fertilized egg is 19 days.  Suppose the incubation times are approximately normally distrubuted with a standard deviation of 1 day.

What is the probability a randoly selected egg hatches in less than 18 days

What is the probability a randomly selected egg hatches takes over 20 days

what is the probabilty that randomly selected fertilized eggs hatches 17 and 19 days?

Would it be unusual for an egg to hatch in less than 17.5days and why?

Answer 1

Because you know the population standard deviation and know the population follows a normal distribution, you can use the z-test.

(A) What is the probability a randoly selected egg hatches in less than 18 days?

Construct your test statistic:

P(< 18 days)

z< (x-μ)/σ

z < (18-19)/1

z<-1

Using a probability table (remember- this is one-sided), this corresponds to p = 0.1587.

(B) What is the probability a randomly selected egg hatches takes over 20 days?

Construct your test statistic:

P(> 20 days)

z > (20-19)/1

z<1

Using a probability table (remember- this is one-sided), this corresponds to p = 0.1587.

(C) What is the probabilty that randomly selected fertilized eggs hatches 17 and 19 days?

Construct your test statistic:

P(17<x< 19 days)

(17-19)/1 <z  (19-19)/1

-2<z<0

This means we calculate

P(z<0) - P(z<-2)

= 0.5000 - 0.0228

= 0.4772

 

(D) Would it be unusual for an egg to hatch in less than 17.5 days and why?

Ugh. It depends what you mean by unusual.

t = 17.5 days corresponds to a z score of 

 

z = (17.5-19)/1 = -1.5

which means a probability of 0.0668.

If you use a threshold of α=0.10 (10%) as "unusual", then yes, it is. However, it is more typical to use an α = 0.05, in which case this doesn't count as unusual.

Answer 2

<X> = 19 days

s = 1 day

 

Z= (x-<x>)/s

 

1. What is the probability,p, that x<18 or Z<-1 from a table of the cumluative normal distribution we find p=.155

2.Here Z>1which is the same as Z<-1

3. p=0.053990967*0.39894228 = 0.021539279 i.e.. probability Z=-2 times probability Z=0

4. Here Z=(17.5-19)/1 = -1.5 the probability of Z being less than -1.5 is about 0.066 so this would happen only about 7 times out of every hundred eggs not very often