Question

3. A factory produces whiteboard markers. The production process can be modelling by normal distribution with a mean length of 11 cm and a standard deviation of 0.25 cm. (a) What is the probability that a randomly selected whiteboard marker has a length longer than 10.5 cm? 100 whiteboard markers are randomly selected for quality checking. (2 marks) (b) What are the mean and standard deviation of the sample mean length? (2 marks) (c) What is the probability that the sample mean length will be between 10.99 cm and 11.01 cm? (3 marks) (d) If 99.8% of the sample means are more than a specific length Y, find Y. (3 marks) (On doing Q3, please refer to your normal distribution table in your notes)

Answer 1

Solution:

Given that , X follows normal distribution with

$\mu$ = 11

$\sigma$ = 0.25

a)

P(X is longer than 10.5)

= P(X > 10.5 )

=  P[(X - $\mu$ )/$\sigma$ > (10.5 - 11)/0.25]

= P[Z > -2.00 ]

=  1 - P[Z < -2.00]

= 1 - 0.0228

= 0.9772

P(X is longer than 10.5) =  0.9772

b)

Suppose , sample of size n = 100 is taken from given population.

Let $\bar x$ be the mean of sample.

The sampling distribution of the $\bar x$ is approximately normal with

Mean $\mu_{\bar x}$ = $\mu$ = 11

SD $\sigma_{\bar x}$ =    $\sigma/\sqrt{n}$   = 0.25/$\sqrt{}$​100= 0.025

c)

P(The sample mean is between 10.99 and 11.01)

= P(10.99 < $\bar x$ < 11.01)

= P($\bar x$ < 11.01) - P($\bar x$ < 10.99)

= P[($\bar x$ - $\mu_{\bar x}$ )/$\sigma_{\bar x}$ < (11.01 - 11)/0.025] - P[($\bar x$ - $\mu_{\bar x}$ )/$\sigma_{\bar x}$ < (10.99- 11)/0.025]

= P[Z < 0.40] - P[Z < -0.40]

=  0.6554 - 0.3446(use z table)

= 0.3108

P(The sample mean is between 10.99 and 11.01) = 0.3108

d)

Given that ,

P($\bar x$ > Y ) = 99.8%

P($\bar x$ > Y ) = 0.998

P($\bar x$ < Y ) = 1 - 0.998

P($\bar x$ < Y ) = 0.0020

For z value , P(Z < z) = 0.0020

But from z table , P(Z < -2.88) = 0.0020

So , z = -2.88

Now , using z score formula ,

Y = $\mu_{\bar x}$ + (z * $\sigma_{\bar x}$ ) = 11 + (-2.88 * 0.025) = 10.928

Y = 10.928