Question

I. In a shipment of 300 connecting rods, the mean tensile strength is found to be 45 kpsi and has a standard deviation of 5 kpsi. (a) Assuming a normal distribution, how many rods can be expected to have a strength less than 39.4 kpsi? (b) How many are expected to have a strength between 39.4 and 60 kpsi?

Answer 1

Answer:

a) Between 39 and 40 rods can be expected to have a strength less than 39.4 kpsi.

b) 260 rods are expected to have a strength between 39.4 and 60 kpsi

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 45, \sigma = 5$

(a) Assuming a normal distribution, how many rods can be expected to have a strength less than 39.4 kpsi?

The percentage of rods with a stength less than 39.4 is the pvalue of Z when X = 39.4. So

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

$Z = \frac{39.4 - 45}{5}$

$Z = -1.12$

$Z = -1.12$ has a pvalue of 0.1314

13.14% of rods have a strength less than 39.4 kpsi.

Out of 300

0.1314*300 = 39.42

Between 39 and 40 rods can be expected to have a strength less than 39.4 kpsi.

(b) How many are expected to have a strength between 39.4 and 60 kpsi?

The percentage of rods with a stength in this interval is the pvalue of Z when X = 60 subtracted by the pvalue of Z when X = 39.4. So

X = 60

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

$Z = \frac{60 - 45}{5}$

$Z = 3$

$Z = 3$ has a pvalue of 0.9987

X = 39.4

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

$Z = \frac{39.4 - 45}{5}$

$Z = -1.12$

$Z = -1.12$ has a pvalue of 0.1314

0.9987 - 0.1314 = 0.8673

86.73% of the rods are expected to have a strength between 39.4 and 60 kpsi

Out of 300

0.8673*300 = 260

260 rods are expected to have a strength between 39.4 and 60 kpsi