Question: Question 1: Answer all 5 boxes below: Sharon's mp music files have a mean size of 4.0 megabytes and a standard deviation of 1.8 megabytes.

Question 1: Answer all 5 boxes below:

Question 1: Answer all 5 boxes below: Sharon's mp music files havea mean size of 4.0 megabytes and a standard deviation of 1.8megabytes. She wants to create a mix of 10 songs for afriend Let the random variable T = the total size (in megabytes)for 10 randomly selected songs from Sharon's computer. So, T = 10(filesize). Find the expected value (mean) of T. O 4.0 O 10.0O 14.0 O 40.0Sharon's mp music files have a mean size of

Sharon's mp music files have a mean size of 4.0 megabytes and a standard deviation of 1.8 megabytes. She wants to create a mix of 10 songs for a friend Let the random variable T = the total size (in megabytes) for 10 randomly selected songs from Sharon's computer. So, T = 10(file size). Find the expected value (mean) of T. O 4.0 O 10.0 O 14.0 O 40.0Sharon's mp music files have a mean size of 4.0 megabytes and a standard deviation of 1.8 megabytes. She wants to create a mix of 10 songs for a friend Let the random variable T = the total size (in megabytes) for 10 randomly selected songs from Sharon's computer. So, T = 10(file size). Find the standard deviation of T. O 1.8 O 3.24 O 18.0 O 32.4Sharon's mp music tiles have a mean size ofii megabytes and a standard deviation of 1.3 megabytes. She wants to create a mix of 10 songs for a Tnend. Let the random variable T = the total size (in megabytes} for 1D randome selected songs from Sharon's computer. So. T = iile size}. Typically, the formula 1.0?(total size] 2.5 provides a good estimate of the length of a song in minutes If M = 1.0?T 2.5. what are the mean and standard deviation of M? \"M24413, oM=1_? O \"M2423, oM=19_2 O \"M24413, oM=19_2 O 0 [1512423, UM=1B.?5 A small store keeps track of the number X of customers that make a purchase during the first hour that the store is open each day. Based on the records, X has the following probability distribution. X 0 1 2 3 4 P(X) 0.1 0.1 0.1 0.1 0.6 The mean number of customers that make a purchase during the first hour that the store is open is O 2.0 O 2.5 O 3.0 O 4.0A small store keeps track of the number X of customers that make a purchase during the first hour that the store is open each day. Based on the records, X has the following probability distribution. X 0 1 2 3 4 P(X) 0.1 0.1 0.1 0.1 0.6 The standard deviation of the number of customers that make a purchase during the first hour that the store is open is O 0.2 O 0.3 O 1.4 O 2.0Mrs. Doe gave her students a 10-question multiple choice quiz. Let Y represent the number of questions that a randomly selected student in the class answered correctly. The computer output below gives information about the probability distribution of Y. N Mean Median StDev Min Max Q1 Qa 25 9.2 7.4 0.9 5 10 6 9 To determine each student's grade on the quiz (out of 100), Mrs. Doe will multiply each student's number of correct answers by 6 and then add 40 Let G = the grade of a randomly chosen student in the class. So, G = 6Y + 40. (Round to the tenths place.) Find the mean of G. Find the standard deviation of G.The manager of a children's puppet theater has determined that the number of tickets he sells for a Saturday afternoon show is a random variable (T) a mean of 32.7 and a standard deviation of 4.9 tickets. The tickets sell for $12. Let M = the money he collects from tickets on a random Saturday. (M = 12T) What are the mean and standard deviation of M? (Round to hundredths place. Do not use units in answer.) . Mean of M = . Standard Deviation of M = It costs $125 for the manager to put on each show. Let P = profit from a random Saturday's show. So P = M - 125. What are the mean and standard deviation of P? (Round to hundredths place. Do not use units in answer.) . Mean of P = . Standard Deviation of P = Assuming the probability distribution of P (profit) is normally distributed, what is the probability the profit for the show is greater than $200? (Round to hundredths place. Write as a decimal - not percentage.) . P(profit > 200) =

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