Suppose you are working with a data set that is normally distributed, with a mean of 200 and a standard deviation of 47. Determine the value of x from the following information. a. 60% of the values are greater than x. b. x is below 17% of the values. c. 22% of the values are less than x. d. x is greater than 55% of the values.

(a)  188.09

(b)  244.84

(c)  163.71

(d) 205.91

Step-by-step explanation:

x = z*σ + μ (1)

x= Unknown value

z = z-score

σ = Standard deviation

μ = mean

(a) At 60% greater than x, 60% of the area is in the right tail.

The Find z-score of  the marked off region is given :

invNorm(1-0.6) = -0.2533

Substituting into (1) we have:

x = -0.2533*47 + 200

  = -11.9051 + 200

  = 188.09

(b) At  x less than 17%, 17% of the area is in the right tail.

The Find z-score of  the marked off region is given :

z score = invNorm(1-0.17)

            = invNorm(0.83)

            =  0.9542

Substituting into (1) we have:

x = 0.9542*47 + 200

  = 44.84 + 200

  = 244.84

(c) At 22% lesser than x, 22% of the area is in the left tail.

The Find z-score of  the marked off region is given :

z score = invNorm(0.22)

            =  -0.7722

Substituting into (1) we have:

x = -0.7722*47 + 200

  = 163.71

(d) At  x greater than 55%, 55% of the area is in the left tail.

The Find z-score of  the marked off region is given :

z score = invNorm(0.55)

            =  0.1257

Substituting into (1) we have:

x = 0.1257*47 + 200

  = 5.91 + 200

  = 205.91

21/58.

hope to be marked brainliest

no this is a nonlinear negative