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3 years ago

12

In a certain region, the average movie theater ticket prices in 2009 was $7.70, in 2008 it was $7.36. Find the increase in avera

ge movie theater ticket price from 2008 to 2009

1 answer:

nordsb [41]3 years ago

8 0

Answer:

The increase price of movie ticket is $ 0.34 or 4.619 %

Step-by-step explanation:

Given as :

The average movie theater price in 2009 = $ 7.70

The average movie theater price in 2008 = $ 7.36

So, let The increase in price from 2008 to 2009 = $ 7.70 - $ 7.36

I.e = $ 0.34

Now in percentage increase

Percentage increase = $\frac{price in 2009 - price in 2008}{price in 2008}$ × 100

Or, Percentage increase = $\frac{$ 7.70 - $ 7.36}{$ 7.36}$ × 100

Or, Percentage increase = $\frac{$ 0.34}{$ 7.36}$ × 100

∴ Percentage increase = 4.619 %

Hence The increase price of movie ticket is $ 0.34 or 4.619 % Answer

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2 years ago

The amount of time all students in a very large undergraduate statistics course take to complete an examination is distributed c

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Answer:

a) The mean is $\mu = 60$

b) The standard deviation is $\sigma = 9$

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

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.

The probability a student selected at random takes at least 55.50 minutes to complete the examination equals 0.6915.

This means that when X = 55.5, Z has a pvalue of 1 - 0.6915 = 0.3085. This means that when $X = 55.5, Z = -0.5$

So

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

$-0.5 = \frac{55.5 - \mu}{\sigma}$

$-0.5\sigma = 55.5 - \mu$

$\mu = 55.5 + 0.5\sigma$

The probability a student selected at random takes no more than 71.52 minutes to complete the examination equals 0.8997.

This means that when X = 71.52, Z has a pvalue of 0.8997. This means that when $X = 71.52, Z = 1.28$

So

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

$1.28 = \frac{71.52 - \mu}{\sigma}$

$1.28\sigma = 71.52 - \mu$

$\mu = 71.52 - 1.28\sigma$

Since we also have that $\mu = 55.5 + 0.5\sigma$

$55.5 + 0.5\sigma = 71.52 - 1.28\sigma$

$1.78\sigma = 71.52 - 55.5$

$\sigma = \frac{(71.52 - 55.5)}{1.78}$

$\sigma = 9$

$\mu = 55.5 + 0.5\sigma = 55.5 + 0.5*9 = 55.5 + 4.5 = 60$

Question

The mean is $\mu = 60$

The standard deviation is $\sigma = 9$

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