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 normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $n\mu$ and the standard error is $s = \sigma\sqrt{n}$

In this problem:

Mean of $0.47, standard deviation $0.15, hence $\mu = 0.47, \sigma = 0.15$

30 instances, hence $n\mu = 30(0.47) = 14.1, s = 0.15\sqrt{30} = 0.8216$

The probability is 1 subtracted by the p-value of Z when X = 15, hence:

$Z = \frac{X - \mu}{\sigma}$

Considering the n instances:

$Z = \frac{X - n\mu}{s}$

$Z = \frac{15 - 14.1}{0.8216}$

$Z = 1.1$

$Z = 1.1$ 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

6 0

2 years ago

What formula can be used to describe the sequence below? -8,-5,-2,1,4...

ikadub [295]

This is just an arithmetic sequence because each term is a constant difference from the preceding term, called the common difference.

Any arithmetic sequence can be expressed as:

a(n)=a+d(n-1) where a=initial term, d=common difference, n=term number

In this case a=-8 and d=3 so

a(n)=-8+3(n-1) which you can simplify if you'd like

a(n)=-8+3n-3

a(n)=3n-11

7 0

3 years ago

Read 2 more answers

3. During the summer Noah plays on a travel basketball team. His points

alexdok [17]

Answer:

sorry i dont kno

Step-by-step explanation:

5 0

3 years ago