Bob's average golf score at his local course is 93.6 from a random sample of 34 rounds of golf. Assume that the population standard deviation for his golf score is 4.2. The 90% confidence interval around this sample mean is:. a. (92.4, 94.8).
b. (89.0, 98.3).
c. (90.5, 96.7).
d. (93.2, 94.0).

a. (92.4, 94.8)

Step-by-step explanation:

Given that:

Mean (m) = 93.6

Standard deviation (σ) = 4.2

Sample size (n) = 34

α = 90%

Confidence interval around sample mean :

Mean ± Zstatistic * s/√n

Z0.90 = 1.645 (Z probability calculator)

s / √n = 4.2 / √34 = 0.7202940

Lower limit = 93.6 - (1.645 * 0.7202940) = 92.41511637 = 92.4

Upper limit = 93.6 + (1.645 * 0.7202940) = 94.78488363 = 94.8

step-by-step explanation:

107 units

javeir earned 56 dollors more than kyle so kyle and javier had 14w but javier made 56 dollors more so 14w plus 56 equals 70w so kyle has 14w

step-by-step explanation: hope i plz give me brainliest with the points so this is not constant cause it says last week so its clearly not every week