A scientist studying water quality measures the lead level in parts per billion (ppb) at each of 49 randomly chosen locations along a water line. Suppose that the lead levels across all the locations on this line are strongly skewed to the right with a mean of ppb17 and a standard deviation of 14ppb. Assume that the measurements in the sample are independent. Required:
What is the probability that the mean lead level from the sample of 49 measurements T is less than 15 ppb?

The probability that the mean lead level from the sample of 49 measurements T is less than 15 ppb

P(x⁻< 15) = 0.1587

Step-by-step explanation:

Step(i):-

Given that the size of the sample 'n' =49

Mean of the Population = 17ppb

The standard deviation of the population = 14ppb

Let 'X' be the random variable in a normal distribution

Step(ii):-

The probability that the mean lead level from the sample of 49 measurements T is less than 15 ppb

P(x⁻< 15) = P(Z<-1) = 1-P(Z>-1)

                             = 1-(0.5+A(-1))

                             = 0.5 - A(1)

                            = 0.5-0.3413

                            = 0.1587

Final -

The probability that the mean lead level from the sample of 49 measurements T is less than 15 ppb

P(x⁻< 15) = 0.1587

y-y1=4(x-x1)

step-by-step explanation:

this is point-slope form since you only have 4 as the slope it goes in place of m

hope this : )

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204