Find the difference 18,354 4,672
1 answer:
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Answer:
a. b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349
Step-by-step explanation:
a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.
b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.
c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842
d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166
e. The z-score related to 6.4 kg is and the z-score related to 7 kg is
, we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194
f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349
Answer:
Yes, he made a reasonable inference by saying about 25 percent of students choose the science.
Step-by-step explanation:
Given that
Number of students for survey = 50
Number of students who choose science = 12
Now
Percentage of student who choose science =
Percentage of student who choose science =
Percentage of student who choose science = 24 %
Now as Mr. Yule said "about 25" percent of students choose the science it can be treated as reasonable approximation.
Now if the number of students in survey become 200, then we can expect 12 out of every 50 students would choose the science. So, in total 48 students would choose the science.
Now lets calculate the percentage of students who choose science.
Percentage of student who choose science =
Percentage of student who choose science =
Percentage of student who choose science = 24 %