Rick’s scores in his last 15 rounds of golf are 72, 80, 75, 84, 78, 72, 75, 68, 81, 74, 79, 70, 77, 83, and 72. If he scores an 80 on each of the next three rounds, which measure of center would increase more, the mean or median?

The median increased more

Step-by-step explanation:

Given that:

Scores on last 15 :

72, 80, 75, 84, 78, 72, 75, 68, 81, 74, 79, 70, 77, 83, 72

Obtaining the range, mean, median and mode from the data above :

Rearranged data:

68, 70, 72, 72, 72, 74, 75, 75, 77, 78, 79, 80, 81, 83, 84

Range = 84 - 64 = 16

Mode = 72 - frequency of 3)

Mean = ΣX / n where n = sample size

Mean = 1140 / 15 = 76

Median = 0.5(n+1)th term = 0.5(16)th term =

8th term = 75

A score of 80 on each of next 3 rounds :

72, 80, 75, 84, 78, 72, 75, 68, 81, 74, 79, 70, 77, 83, 72, 80, 80, 80

Rearranged data:

68, 70, 72, 72, 72, 74, 75, 75, 77, 78, 79, 80, 80, 80, 80, 81, 83, 84

Range = 84 - 68 = 16

Mean = ΣX / n where n = sample size

Mean = 1380 / 18 = 76.67

Median =0.5(n+1)th term = 0.5(19)th term =

9.5 = (9th term +. 10th term) / 2 = (77 +78) / 2 = 77.5

you would have to sell 24 bracelets.

step-by-step explanation:

84/3.5 = 24

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;