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.

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

2) P17 = 125.57cm

Step-by-step explanation:

Normal Probability Distribution:

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

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

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

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

has a pvalue of 0.5871

X = 258.5

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

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.

P17 = 125.57cm

80 miles were driven after   of the gas was used.

Step-by-step explanation:

The car gets 30 miles in one gallon and we have to calculate the miles driven after 2 gallons of gas was used.

We will use unitary method to solve this question.

∵ In one gallon of gas car runs = 30 miles

∴ In 8/3 gallons the car will cover the distance = = 8×10 = 80 miles

Therefore the car will be driven 80 miles with of gas.