Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.5. (Round your answers to four decimal places.)(a) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 10 pins is at least 51

0.0174 = 1.74% probability that the sample mean hardness for a random sample of 10 pins is at least 51

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean and standard deviation , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

What is the probability that the sample mean hardness for a random sample of 10 pins is at least 51

This is 1 subtracted by the pvalue of Z when X = 51. So

By the Central Limit Theorem

has a pvalue of 0.9826

1 - 0.9826 = 0.0174

0.0174 = 1.74% probability that the sample mean hardness for a random sample of 10 pins is at least 51

32/80  

91/140

the last one 54 doesn't go into 45. 54/100 ?

step-by-step explanation:

yes because they have the same amount of green and blur marbles.

step-by-step explanation: