Answer:
61 tickets
Step-by-step explanation:
It is given that,
The total profit from the Spanish club fund-raiser varies directly with the number of dinner tickets sold.
If the sale of 22 tickets results in a profit of $253.
Profit in 1 ticket =
Let x tickets are sold in order to reach the fund-raising goal of $700.
or
= 61 tickets
Hence, 61 tickets must be sold to reach the fund-raising goal of $700.
Answer:
a) 0.0016 = 0.16% probability that the sample mean is at least $30.00.
b) 0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00
c) 90% of sample means will occur between $26.1 and $28.9.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
a. What is the likelihood the sample mean is at least $30.00?
This is 1 subtracted by the pvalue of Z when X = 30. So
By the Central Limit Theorem, we have that:
has a pvalue of 0.9984
1 - 0.9984 = 0.0016
0.0016 = 0.16% probability that the sample mean is at least $30.00.
b. What is the likelihood the sample mean is greater than $26.50 but less than $30.00?
This is the pvalue of Z when X = 30 subtracted by the pvalue of Z when X = 26.50. So
From a, when X = 30, Z has a pvalue of 0.9984
When X = 26.5
has a pvalue of 0.1190
0.9984 - 0.1190 = 0.8794
0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00.
c. Within what limits will 90 percent of the sample means occur?
Between the 50 - (90/2) = 5th percentile and the 50 + (90/2) = 95th percentile, that is, Z between -1.645 and Z = 1.645
Lower bound:
Upper Bound:
90% of sample means will occur between $26.1 and $28.9.