Question

Specifications for an aircraft bolt require that the ultimate ten- sile strength be at least 18 kN. It is known that 10% of the bolts have strengths less than 18 . 3 kN and that 5% of the bolts have strengths greater than 19 . 76 kN. It is also known that the strengths of these bolts are normally distributed.


a. Find the mean and standard deviation of the strengths.

b. What proportion of the bolts meet the strength specification?

Answer 1

Answer:

a) $P(X <18.3)= 0.1$   (1)

$P(X>19.76) = 0.05$   (2)

Using the condition (1) and the z score we have:

$P(X <18.3)=P(z<\frac{18.3 -\mu}{\sigma})= 0.1$

We can find a z score that accumulates 0.1 of the area on the left and we got z = -1.28, and we can rewrite:

$-1.28 = \frac{18.3 -\mu}{\sigma}$   (1)

And similarly for the other condition we have:

A value that accumulates 0.95 of the area on the left is z=1.64 and we have:

$1.64 = \frac{19.76 -\mu}{\sigma}$   (2)

Using equation (1) we got:

$-1.28 \sigma -18.3 = -\mu$

$\mu = 1.28\sigma +18.3$   (3)

And replacing this into the equation (2) we got:

$1.64 \sigma = 19.76 - \mu$

$1.64 \sigma =19.76 -1.28 \sigma -18.3$

$2.92 \sigma = 1.46$

$\sigma = 0.5$

And for the mean we got:

$\mu = 1.28*0.5 +18.3 = 18.94$

b) $P(X>18) = 1-P(X<18 ) =1-P(Z< \frac{18-18.94}{0.5}) = 1-P(Z<-1.88) =1- 0.03 = 0.97$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the tensile strength of a population, and for this case we know the distribution for X is given by:

$X \sim N(\mu,\sigma)$  

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

We know the following conditions:

$P(X <18.3)= 0.1$   (1)

$P(X>19.76) = 0.05$   (2)

Using the condition (1) and the z score we have:

$P(X <18.3)=P(z<\frac{18.3 -\mu}{\sigma})= 0.1$

We can find a z score that accumulate 0.1 of the area on the left and we got z = -1.28, and we can rewrite:

$-1.28 = \frac{18.3 -\mu}{\sigma}$   (1)

And similarly for the other condition we have:

A value that accumulate 0.95 of the area on the left is z=1.64 and we have:

$1.64 = \frac{19.76 -\mu}{\sigma}$   (2)

Using equation (1) we got:

$-1.28 \sigma -18.3 = -\mu$

$\mu = 1.28\sigma +18.3$   (3)

And replacing this into the equation (2) we got:

$1.64 \sigma = 19.76 - \mu$

$1.64 \sigma =19.76 -1.28 \sigma -18.3$

$2.92 \sigma = 1.46$

$\sigma = 0.5$

And for the mean we got:

$\mu = 1.28*0.5 +18.3 = 18.94$

Part b

For this case we want this probability:

$P(X>18)$

And using the z score we got:

$P(X>18) = 1-P(X<18 ) =1-P(Z< \frac{18-18.94}{0.5}) = 1-P(Z<-1.88) =1- 0.03 = 0.97$