Respuesta :
The probability that a randomly selected bill will be at least $39.10 is 0.03216 percent.
According to the question, given that
The average daily dinner tab at a nearby restaurant has a normal distribution with a standard deviation of $6 and a mean of $28.
Let X = daily dinner bills in a local restaurant
So, X ~ N(μ = 28, σ^2 = [tex]6^{2}[/tex] )
The following equations give the normal distribution's z-score probability distribution:
Z = (X - μ) /σ ~ N(0,1)
where, μ = mean amount = $28
σ = standard deviation = $6
The Z-score calculates the deviation of the measure from the mean in standard deviations. We look at the z-score table after determining the Z-score to determine the p-value (area) connected to it. The likelihood that the measure's value is less than X, or the percentile of X, is represented by this p-value.
As a result, = P(X $39.10) is the chance that a randomly chosen bill will be at least $39.10.
P(X ≥ $39.10) = P (X - μ) /σ ≥ [tex]\frac{39.10 - 28}{6}[/tex]) = P(Z ≥ 1.85) = 1 - P(Z < 1.85)
= 1 - 0.96784 = 0.03216
Now, the P(Z x) or P(Z x) is presented in the z table. By examining the value of x = 1.85 in the z table, which has an area of 0.96784, the probability stated above is derived.
Therefore, there is a 0.03216 percent chance that a randomly chosen bill will be at least $39.10.
To learn more about standard deviations visit here : https://brainly.com/question/13905583
#SPJ4
