Normal distribution with a probability less than 0.1?

Question

The question states, "find the value of c where Z has a normal distribution with u=0 and variance=1

P(-2.7 < Z < c) = 0.0252

How do I find the c value for this problem?

Answer 1

Hint:

• $\mathbb P(-2.7 < Z < c) = \mathbb P(Z \lt c) - \mathbb P(Z \le -2.7)$

• You can find $\mathbb P(Z \le -2.7)$

• and then find $c$ such that $\mathbb P(Z \lt c) = \mathbb P(Z \le -2.7) + 0.0252$

Answer 2

Thank you for that help. I have learnt to use a normal distribution table. I did the following working, bearing in mind your suggestion and got the question right:

P (Z < c) - 1 + 0.9965 = 0.0252 = P (Z < c) - 0.0035 = 0.0252 Therefore P(Z < c) = 0.0287 If d = (1 - c), Then P(Z < c) = 0.9713 Which means that d = 1.90

Therefore c = -1.90

This was the correct answer in the textbook