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A company’s cereal boxes advertise that 9.65 ounces of cereal are contained in each box. In fact, the amount of cereal in a box follows an approximately normal distribution with mean μ = 9.70 ounces and standard deviation σ = 0.03 ounce. Let x-bar be the sample mean amount of cereal in 5 randomly selected boxes.
A company’s cereal boxes advertise that 9.65 ounces of cereal are contained in each box. In fact, the amount of cereal in a box follows an approximately normal distribution with mean μ = 9.70 ounces and standard deviation σ = 0.03 ounce. Let x-bar be the sample mean amount of cereal in 5 randomly selected boxes.Shape: Approximately normal because the population distribution is approximately normal
Center: μ x-bar = 9.70 ounces.
Variability: σ x-bar = 0.013 ounces.
(b) What is the probability that the mean amount of cereal x-bar in 5 randomly selected boxes is at most 9.65?
Round your answer to 5 decimal places.
Theprobabilitythat themeanamount of cereal x-bar in 5 randomly selected boxes is at most 9.65 is 0.0001.How to calculate the probabilityProbabilitysimply means the chance that a particular thing or event will happen. It is the occurence of likely events. It is simply the area of mathematics that deals with the numerical estimates of the chance that an event will occur or that a particular statement is true.Mean μ=9.70Sample size=n=5standarddeviation=σ= 0.03Now in order to use thetablewe have to to figure out z value.z=(x-μ)/σz=(9.65-9.70)/0.0134z=-3.73From theprobabilityindex tables, so P(z≤-3.73)=0.0001Learn more aboutprobabilityonbrainly.com/question/14859497#SPJ1...
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