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.