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The mean preparation fee h&r block charged retail customers last year was $183 (the wall street journal, march 7, 2012). use this price as the population mean and assume the population standard deviation of preparation fees is $50.
a. what is the probability that the mean price for a sample of 30 h&r block retail customers is within $8 of the population mean?
b. what is the probability that the mean price for a sample of 50 h&r block retail customers is within $8 of the population mean?
c. what is the probability that the mean price for a sample of 100

Respuesta :

nobillionaireNobley
Given that the population mean, [tex]\mu=\$183[/tex] and the population standard deviation, [tex]\sigma=\$50[/tex]

Part A:

The probability that the mean price for a sample of 30 h&r block retail customers is within $8 of the population mean is evaluated as follows:

[tex]P(183-8\leq\bar{x}\leq183+8)=P(175\leq\bar{x}\leq191) \\ \\ =P(\bbar{x}\leq191)-P(\bbar{x}\leq175)=P\left(z\leq \frac{191-183}{ \frac{50}{ \sqrt{30} } } \right)-P\left(z\leq \frac{175-183}{ \frac{50}{ \sqrt{30} } } \right) \\ \\ =P(z\leq0.8764)-P(z\leq-0.8764) \\ \\ =P(z\leq0.8764)-[1-P(z\leq0.8764)] \\ \\ =2P(z\leq0.8764)-1=2(0.80958)-1 \\ \\ =1.61916-1=0.61916\approx62\%[/tex]



Part B:

The probability that the mean price for a sample of 50 h&r block retail customers is within $8 of the population mean is evaluated as follows:

[tex]P(183-8\leq\bar{x}\leq183+8)=P(175\leq\bar{x}\leq191) \\ \\ =P(\bbar{x}\leq191)-P(\bbar{x}\leq175)=P\left(z\leq \frac{191-183}{ \frac{50}{ \sqrt{50} } } \right)-P\left(z\leq \frac{175-183}{ \frac{50}{ \sqrt{50} } } \right) \\ \\ =P(z\leq1.131)-P(z\leq-1.131) \\ \\ =P(z\leq1.131)-[1-P(z\leq1.131)] \\ \\ =2P(z\leq1.131)-1=2(0.87105)-1 \\ \\ =1.7421-1=0.7421\approx74\%[/tex]



Part C:

The probability that the mean price for a sample of 50 h&r block retail customers is within $8 of the population mean is evaluated as follows:

[tex]P(183-8\leq\bar{x}\leq183+8)=P(175\leq\bar{x}\leq191) \\ \\ =P(\bbar{x}\leq191)-P(\bbar{x}\leq175)=P\left(z\leq \frac{191-183}{ \frac{50}{ \sqrt{100} } } \right)-P\left(z\leq \frac{175-183}{ \frac{50}{ \sqrt{100} } } \right) \\ \\ =P(z\leq1.6)-P(z\leq-1.6) \\ \\ =P(z\leq1.6)-[1-P(z\leq1.6)] \\ \\ =2P(z\leq1.6)-1=2(0.9452)-1 \\ \\ =1.8904-1=0.8904\approx89\%[/tex]

a) 62%

b) 74%

c) 89%

Step-by-step explanation:

Given :

Mean, [tex]\mu = 183[/tex]

Standard Deviation, [tex]\sigma = 50[/tex]

Solution :

a) [tex]\rm P(183-8\leq \bar{x} \leq 183+8 ) = P(175\leq \bar{x} \leq 191)[/tex]

[tex]= \rm P(x\leq 195) - P(x\leq 175)[/tex]

[tex]\rm = P(z \leq \dfrac{191-183}{\dfrac{50}{\sqrt{30} } }) - P(z \leq \dfrac{175-183}{\dfrac{50}{\sqrt{30} } })[/tex]

[tex]=\rm P(z\leq 0.8764) - P(z\leq -0.8764)[/tex]

[tex]=\rm P(z\leq 0.8764) - (1-P(z\leq 0.8764))[/tex]

[tex]\rm =2(P(z\leq 0.8764))-1=2(0.80958)-1[/tex]

= 0.61916

= 62 % (Approx)

b) [tex]\rm P(183-8\leq \bar{x} \leq 183+8 ) = P(175\leq \bar{x} \leq 191)[/tex]

[tex]= \rm P(x\leq 195) - P(x\leq 175)[/tex]

[tex]\rm = P(z \leq \dfrac{191-183}{\dfrac{50}{\sqrt{50} } }) - P(z \leq \dfrac{175-183}{\dfrac{50}{\sqrt{50} } })[/tex]

[tex]=\rm P(z\leq 1.131) - P(z\leq -1.131)[/tex]

[tex]=\rm P(z\leq 1.131) - (1-P(z\leq 1.131))[/tex]

[tex]\rm =2(P(z\leq 1.131))-1=2(0.87105)-1[/tex]

= 0.7421

= 74 % (Approx)

c) [tex]\rm P(183-8\leq \bar{x} \leq 183+8 ) = P(175\leq \bar{x} \leq 191)[/tex]

[tex]= \rm P(x\leq 195) - P(x\leq 175)[/tex]

[tex]\rm = P(z \leq \dfrac{191-183}{\dfrac{50}{\sqrt{100} } }) - P(z \leq \dfrac{175-183}{\dfrac{50}{\sqrt{100} } })[/tex]

[tex]=\rm P(z\leq 1.6) - P(z\leq -1.6)[/tex]

[tex]=\rm P(z\leq 1.6) - (1-P(z\leq 1.6))[/tex]

[tex]\rm=2(P(z\leq 1.6))-1=2(0.9452)-1[/tex]

= 0.8904

= 89 % (Approx)

For more information, refer the link given below

https://brainly.com/question/21586810?referrer=searchResults

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