Answered

Join IDNLearn.com and become part of a knowledge-sharing community that thrives on curiosity. Join our interactive Q&A community and access a wealth of reliable answers to your most pressing questions.

A racing car consumes a mean of 98 gallons of gas per race with the standard deviation of 3 gallons. If 42 racing cars are randomly selected what is the probability that the sample mean will be greater than 98.6 gallons? Round your answer to four decimal places

A Racing Car Consumes A Mean Of 98 Gallons Of Gas Per Race With The Standard Deviation Of 3 Gallons If 42 Racing Cars Are Randomly Selected What Is The Probabil class=

Sagot :

Given:

Mean= 98

Standard Deviation= 3

Sample = 42

Observed value = 98.6

To determine the probability that the sample mean will be greater than 98.6 gallons, we first note the z-score formula:

[tex]\begin{gathered} z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}} \\ where: \\ x=observed\text{ value} \\ \mu=mean \\ \sigma=standard\text{ deviation} \\ n=sample \end{gathered}[/tex]

Next, we plug in what we know:

[tex]\begin{gathered} z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}} \\ z=\frac{98.6-98}{\frac{3}{\sqrt{42}}} \\ Calculate \\ z=1.2961 \end{gathered}[/tex]

Then, we also note that:

[tex]P(x>98.6)=P(z>1.2961)=0.0975[/tex]

Therefore, the probability that the sample mean will be greater than 98.6 gallons is: 0.0975

Your presence in our community is highly appreciated. Keep sharing your insights and solutions. Together, we can build a rich and valuable knowledge resource for everyone. Accurate answers are just a click away at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.