Answer:
560 students
Step-by-step explanation:
400x1.4=560
Answer:
B!
Step-by-step explanation:
Answer:
X/2 = 61.2
X=122.4
Step-by-step explanation:
Ok so we know that a number was halved so we can use x and equaled to 61.2 so we can do x/2 = 61.2
Less than 90° = acute
more than 90° = obtuse
equal to 180° = straight
more than 360° = reflex
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.
<h3>Normal Probability Distribution</h3>
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
is
.
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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