Question

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

Answer 1

Answer:

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 $\mu$ and standard deviation $\sigma$, the z-score 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 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

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

$Z = \frac{259 - 258.5}{2.3}$

$Z = 0.22$

$Z = 0.22$ has a pvalue of 0.5871

X = 258.5

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

$Z = \frac{258.5 - 258.5}{2.3}$

$Z = 0$

$Z = 0$ 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.

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

$-0.954 = \frac{X - 127.1}{1.6}$

$X - 127.1 = -0.954*1.6$

$X = 125.57$

P17 = 125.57cm