33. A labourer is engaged for 20 days on the condition that he will receive 120 for each day he
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Using the normal distribution, it is found that there is a 0.1357 = 13.57% probability that the total amount paid for these second movies will exceed $15.00.
In a <em>normal distribution</em> with mean and standard deviation
, the z-score of a measure X is given by:
- It 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, which is the percentile of X.
- For n instances of a normal variable, the mean is
and the standard error is
In this problem:
- Mean of $0.47, standard deviation $0.15, hence
- 30 instances, hence
The probability is <u>1 subtracted by the p-value of Z when X = 15</u>, hence:
Considering the n instances:
has a p-value of 0.8643.
1 - 0.8643 = 0.1357.
0.1357 = 13.57% probability that the total amount paid for these second movies will exceed $15.00.
A similar problem is given at brainly.com/question/25769446