Question

The breaking strength of a rivet has a mean value of 10,000 psi and a standard deviation of 500 psi. What is the probability that the sample means strength for a random sample of 15 rivets is between 9900 to 10,200?

Answer 1

Answer:

A manufacturing company claims that the mean breaking strenght of the cables supplied s 1500 with a standard deviation of 100.A sample of

Step-by-step explanation:

Answer 2

Given:

Mean value of breaking strength= μ = 10000

Standard deviation= σ = 500

To find:

The probability that the sample means strength for a random sample of 15 rivets is between 9900 to 10,200

Solution:

n = 15

μ = 10000

σ mean = σ / √n = 500 / √15 = 129.13

P ( 9900 < X < 10200)

= P $[\frac{9900-10000}{129.13} < \frac{X-\mu mean}{\sigma} , \frac{P (10200-10000)}{129.13}]$

= P [ -0.771 < Z < 1.54]

= P ( Z < 1.54) - P (Z < -0.771)

Using Z table,

Probability = 0.938 - 0.220

= 0.718

Hence, the probability is 0.718