Lamar wants to determine the movie preferences
2 answers:
Answer:
Dramas Comedies
Males 9 31
Females 36 24
Step-by-step explanation:
It is given that Lamar surveyed 40 males and 60 females about whether they prefer dramas or comedies.
Total males = 40
Total females = 60
Total people surveyed = 40 + 60 = 100
It is given that 55% of surveyed people preferred comedies.
People preferred comedies =
It is given that males who preferred comedies = 31
Males who preferred drama is the difference of number of males and males who preferred comedies.
Males, drama = 40 - 31 = 9
Females who preferred comedies is the difference of people who preferred comedies and males who preferred comedies.
Females, comedies = 55 - 31 = 24
Females who preferred drama is the difference of number of females and females who preferred comedies.
Females, drama = 60 - 24 = 36
Therefore the table can be filled as
Dramas Comedies
Males 9 31
Females 36 24
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Answer:
The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
What is the minimum weight for a passenger who outweighs at least 90% of the other passengers?
90th percentile
The 90th percentile is X when Z has a pvalue of 0.9. So it is X when Z = 1.28. So
The minimum weight for a passenger who outweighs at least 90% of the other passengers is 203.16 pounds