NEED HELP PLEASE ASAP! please don’t waste my time i need this grade.
1 answer:
Hmm I think the answer is 95%
1 standard deviation is 13
To find out the range of scores that fit 1 standard deviation from the mean, you would have to take away and add 13 from the mean (133) this gives a range of 120-146 which isn’t right because you’re looking for 107-159.
Next, look for 2 standard deviations from the mean which is just 13 x 2=26.
To find the range of scores that are 2 standard deviations from the mean take away and add 26 to the mean (133) which gives a range of 107-159. This is the correct answer.
2 standard deviations represents 95% of the data. That’s why I think the answer is 95%.
1 standard deviation is 13
To find out the range of scores that fit 1 standard deviation from the mean, you would have to take away and add 13 from the mean (133) this gives a range of 120-146 which isn’t right because you’re looking for 107-159.
Next, look for 2 standard deviations from the mean which is just 13 x 2=26.
To find the range of scores that are 2 standard deviations from the mean take away and add 26 to the mean (133) which gives a range of 107-159. This is the correct answer.
2 standard deviations represents 95% of the data. That’s why I think the answer is 95%.
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Answer:
a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.
b) 0.996 is the probability that more than half of the vehicles carry just one person.
Step-by-step explanation:
We are given the following information:
A) Binomial distribution
We treat vehicle on road with one passenger as a success.
P(success) = 64% = 0.64
Then the number of vehicles follows a binomial distribution, where
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 10
We have to evaluate:
0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.
B) By normal approximation
Sample size, n = 92
p = 0.64
We have to evaluate the probability that more than 47 cars carry just one person.
After continuity correction, we will evaluate
Calculation the value from standard normal z table, we have,
0.996 is the probability that more than half out of 92 vehicles carry just one person.