Question

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.2.a. If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 9 pins is at least 51?b. Without assuming population normality, what is the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51?

Answer 1

Answer:

a

 $P(\= X \ge 51 ) =0.0062$

b

$P(\= X \ge 51 ) = 0$

Step-by-step explanation:

From the question we are told that

The mean value is $\mu = 50$

The standard deviation is  $\sigma = 1.2$

Considering question a

The sample size is  n = 9

Generally the standard error of the mean is mathematically represented as

      $\sigma_x = \frac{\sigma }{\sqrt{n} }$

=>   $\sigma_x = \frac{ 1.2 }{\sqrt{9} }$

=>  $\sigma_x = 0.4$

Generally the probability that the sample mean hardness for a random sample of 9 pins is at least 51 is mathematically represented as

      $P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_{x}} \ge \frac{51 - 50 }{0.4 } )$

$\frac{\= X -\mu}{\sigma }  =  Z (The  \ standardized \  value\  of  \ \= X )$

     $P(\= X \ge 51 ) = P( Z \ge 2.5 )$

=>   $P(\= X \ge 51 ) =1- P( Z < 2.5 )$

From the z table  the area under the normal curve to the left corresponding to  2.5  is

    $P( Z < 2.5 ) = 0.99379$

=> $P(\= X \ge 51 ) =1-0.99379$

=> $P(\= X \ge 51 ) =0.0062$

Considering question b

The sample size is  n = 40

   Generally the standard error of the mean is mathematically represented as

      $\sigma_x = \frac{\sigma }{\sqrt{n} }$

=>   $\sigma_x = \frac{ 1.2 }{\sqrt{40} }$

=>  $\sigma_x = 0.1897$

Generally the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51 is mathematically represented as  

       $P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_x} \ge \frac{51 - 50 }{0.1897 } )$

=> $P(\= X \ge 51 ) = P(Z \ge 5.2715 )$

=>  $P(\= X \ge 51 ) = 1- P(Z < 5.2715 )$

From the z table  the area under the normal curve to the left corresponding to  5.2715 and

=>  $P(Z < 5.2715 ) = 1$

So

   $P(\= X \ge 51 ) = 1- 1$

=> $P(\= X \ge 51 ) = 0$

Answer 2

Using the normal distribution and the central limit theorem, we have that there is a:

a) 0.0062 = 0.62% probability that the sample mean hardness for a random sample of 9 pins is at least 51.

b) 0% approximate probability that the sample mean hardness for a random sample of 40 pins is at least 51.

In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  
• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
• By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

In this problem:

• Mean of 50, hence $\mu = 50$
• Standard deviation of 1.2, hence $\sigma = 1.2$
• Sample of 9, hence $n = 9, s = \frac{1.2}{\sqrt{9}} = 0.4$.

Item a:

The probability is 1 subtracted by the p-value of Z when X = 51, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{51 - 50}{0.4}$

$Z = 2.5$

$Z = 2.5$ has a p-value of 0.9938.

1 - 0.9938 = 0.0062

There is a 0.0062 = 0.62% probability that the sample mean hardness for a random sample of 9 pins is at least 51.

Item b:

Even though the distribution is not normal, the sample size is of $n = 40 > 30$, hence the Central Limit Theorem can be applied.

$s = \frac{1.2}{\sqrt{40}} = 0.19$

$Z = \frac{X - \mu}{s}$

$Z = \frac{51 - 50}{0.19}$

$Z = 5.26$

$Z = 5.26$ has a p-value of 1.

1 - 1 = 0

0% approximate probability that the sample mean hardness for a random sample of 40 pins is at least 51.

A similar problem is given at https://brainly.com/question/24663213