Answer: 0.5898
Step-by-step explanation:
Given : J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 .
We assume that,
The probability that .J. Redick makes any given free throw =0.901 (1)
Free throws are independent.
So it is a binomial distribution .
Using binomial probability formula, the probability of getting success in x trials :
, where n= total trials
p= probability of getting in each trial.
Let x be binomial variable that represents the number of a=makes.
n= 14
p= 0.901 (from (1))
The probability that he makes at least 13 of them will be :-
![=^{14}C_{13}(0.901)^{13}(1-0.901)^1+^{14}C_{14}(0.901)^{14}(1-0.901)^0\\\\=(14)(0.901)^{13}(0.099)+(1)(0.901)^{14}\ \ [\because\ ^nC_n=1\ \&\ ^nC_{n-1}=n ]\\\\\approx0.3574+0.2324=0.5898](https://tex.z-dn.net/?f=%3D%5E%7B14%7DC_%7B13%7D%280.901%29%5E%7B13%7D%281-0.901%29%5E1%2B%5E%7B14%7DC_%7B14%7D%280.901%29%5E%7B14%7D%281-0.901%29%5E0%5C%5C%5C%5C%3D%2814%29%280.901%29%5E%7B13%7D%280.099%29%2B%281%29%280.901%29%5E%7B14%7D%5C%20%5C%20%5B%5Cbecause%5C%20%5EnC_n%3D1%5C%20%5C%26%5C%20%5EnC_%7Bn-1%7D%3Dn%20%5D%5C%5C%5C%5C%5Capprox0.3574%2B0.2324%3D0.5898)
∴ The required probability = 0.5898
Answer:
Part A
Please see attached the required stem and leaf plot
For the stem and leaf plot, the nonsplit system is used because of clarity for analysis
Part B:
From the shape of the stem and leaf plot we have that there is an average increase of pulse rate of 20 pulses in all the 19 students after the exercise
The shape of the plot is relatively the same for the before and after exercise save for the decrease in the third to the last row by one and the increase in the second to the last roe by one student
The spread remained relatively constant in both cases with the most being in the 60s range having 7 students in the before exercise and the 80s range having 8 students in the after exercise leaf plot.
Step-by-step explanation:
The given data are;
67
87
67 88
67 89
68 89
71 91
72 93
72 93
75 95
77 96
77 97
79 98
81 98
85 101
87 105
87 105
91 119
97 125
103 125
121 147
Answer:
The standard deviation of the number of rushing yards for the running backs that season is 350.
Step-by-step explanation:
Consider the provided information.
The mean number of rushing yards for the running backs that season is 790 yards. One running back had 1,637 rushing yards for the season, which is 2.42 standard deviations above the mean number of rushing yards.
Here it is given that mean is 790 and 1637 is 2.42 standard deviations above the mean.
Use the formula: 
Here z is 2.42 and μ is 790, substitute the respective values as shown.
Hence, the standard deviation of the number of rushing yards for the running backs that season is 350.
