Question

Someone please help these questions are due at 11;59 today please help urgent

1)A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 258.5-cm and a standard deviation of 2.3-cm. For shipment, 30 steel rods are bundled together.

Find the probability that the average length of a randomly selected bundle of steel rods is between 258.5-cm and 259-cm.
P(258.5-cm < M < 259-cm) =

Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
2)
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 127.1-cm and a standard deviation of 1.6-cm. For shipment, 6 steel rods are bundled together.

Find P17, which is the average length separating the smallest 17% bundles from the largest 83% bundles.
P17 =
-cm
Enter your answer as a number accurate to 2 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

1) P(258.5-cm < M < 259-cm) = 8.71%.

2) P17 = 125.57cm

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (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 p-value, we get the probability that the value of the measure is greater than X.

Question 1:

Mean of 258.5-cm and a standard deviation of 2.3-cm, which means that
`\mu = 258.5, \sigma = 2.3`

Find the probability that the average length of a randomly selected bundle of steel rods is between 258.5-cm and 259-cm.

This is the pvalue of Z when X = 259 subtracted by the pvalue of Z when X = 258.5.

X = 259

`Z = (X - \mu)/(\sigma)`

`Z = (259 - 258.5)/(2.3)`

`Z = 0.22`

`Z = 0.22` has a pvalue of 0.5871

X = 258.5

`Z = (X - \mu)/(\sigma)`

`Z = (258.5 - 258.5)/(2.3)`

`Z = 0`

`Z = 0` has a pvalue of 0.5

0.5871 - 0.5 = 0.0871

0.0871*100% = 8.71%. So

P(258.5-cm < M < 259-cm) = 8.71%.

Question 2:

Mean of 127.1-cm and a standard deviation of 1.6-cm, which means that
`\mu = 127.1, \sigma = 1.6`

Find P17, which is the average length separating the smallest 17% bundles from the largest 83% bundles.

This is the 17th percentile, which is X when Z has a pvalue of 0.17. So X when Z = -0.954.

`Z = (X - \mu)/(\sigma)`

`-0.954 = (X - 127.1)/(1.6)`

`X - 127.1 = -0.954*1.6`

`X = 125.57`

P17 = 125.57cm