PLEASE HELP!!!!! The table displays the mean name length for seven samples of students. Sample Mean Name Length 1 5.4 2 7.1 3 6.
2 answers:
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Using the normal distribution, we have that:
- The distribution of X is
.
- The distribution of
is
.
- 0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.
- 0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.
The z-score of a measure X of a normally distributed variable with mean and standard deviation
is given by:
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation
.
In this problem, the parameters are given as follows:
Hence:
- The distribution of X is
- The distribution of
The probabilities are the <u>p-value of Z when X = 58 subtracted by the p-value of Z when X = 55</u>, hence, for a single movie:
X = 58:
Z = 0.05.
Z = 0.05 has a p-value of 0.5199.
X = 55:
Z = -0.1.
Z = -0.1 has a p-value of 0.4602.
0.5199 - 0.4602 = 0.0597 = 5.97% probability that a single movie production cost is between 55 and 58 million dollars.
For the sample of 17 movies, we have that:
X = 58:
Z = 0.19.
Z = 0.19 has a p-value of 0.5753.
X = 55:
Z = -0.38.
Z = -0.38 has a p-value of 0.3520.
0.5753 - 0.3520 = 0.2233 = 22.33% probability that the average production cost of 17 movies is between 55 and 58 million dollars. Since the sample size is less than 30, assumption of normality is necessary.
More can be learned about the normal distribution at brainly.com/question/4079902
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