There are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters that are normally distributed with mean 3 cm and standard deviation 0.08 cm. The second machine produces corks with diameters that have a normal distribution with mean 3.04 cm and standard deviation 0.04 cm. Acceptable corks have diameters between 2.9 cm and 3.1 cm. What is the probability that the first machine produces an acceptable cork? (Round your answer to four decimal places.) What is the probability that the second machine produces an acceptable cork? (Round your answer to four decimal places.) Which machine is more likely to produce an acceptable cork? the first machine the second machine

Probability that the first machine produces an acceptable cork is 0.7887

Probability that the second machine produces an acceptable cork is 0.9329.

Step-by-step explanation:

We are given that there are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters that are normally distributed with mean 3 cm and standard deviation 0.08 cm. The second machine produces corks with diameters that have a normal distribution with mean 3.04 cm and standard deviation 0.04 cm.

Acceptable corks have diameters between 2.9 cm and 3.1 cm.

Let = diameter for producing corks using first machine

The z-score probability distribution for normal distribution is given by;

            Z =  ~ N(0,1)

where,  = mean diameter of first machine = 3 cm

             =  standard deviation of first machine = 0.08 cm

So,   ~ Normal()

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the first machine produces an acceptable cork is given by = P(2.9 cm < < 3.1 cm) = P( < 3.1 cm) - P( 2.9 cm)

      P( < 3.1 cm) = P( < ) = P(Z < 1.25) = 0.89435

      P( 2.9 cm) = P(   ) = P(Z -1.25) = 1 - P(Z < 1.25)

                                                              = 1 - 0.89435 = 0.10565                       

{Now, in the z table the P(Z  x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.25 in the z table which has an area of  0.89435.}

Therefore, P(2.9 cm < < 3.1 cm) = 0.89435 - 0.10565 = 0.7887

Let = diameter for producing corks using second machine

The z-score probability distribution for normal distribution is given by;

            Z =  ~ N(0,1)

where,  = mean diameter of second machine = 3.04 cm

             =  standard deviation of second machine = 0.04 cm

So,   ~ Normal()

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the second machine produces an acceptable cork is given by = P(2.9 cm < < 3.1 cm) = P( < 3.1 cm) - P( 2.9 cm)

      P( < 3.1 cm) = P( < ) = P(Z < 1.50) = 0.93319

      P( 2.9 cm) = P(   ) = P(Z -3.50) = 1 - P(Z < 3.50)

                                                                  = 1 - 0.99977 = 0.00023

 {Now, in the z table the P(Z  x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.50 and x = 3.50 in the z table which has an area of  and  respectively.}

Therefore, P(2.9 cm < < 3.1 cm) = 0.93319 - 0.00023 = 0.9329

the answer is 175.4379

step-by-step explanation:

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.