Anthony decided to buy two DVDs that were not included in the deal. His price before tax was $19.99. What was the cost of the tw
1 answer:
Given:
Cost of 2 DVDs before tax = $19.99
Tax rate = 8.25%
To find:
The cost of 2 DVDs after the tax was added.
Solution:
We have,
Cost of 2 DVDs before tax = $19.99
Tax rate = 8.25%
So,
Tax = 8.25% of Cost of 2 DVDs before tax
Now,
Cost of 2 DVDs after tax = Cost of 2 DVDs before tax + Tax
= $19.99 + $1.65
= $21.64
Therefore, the cost of 2 DVDs after the tax is $21.64.
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Given Information:
Mean SAT score = μ = 1500
Standard deviation of SAT score = σ = 3 00
Required Information:
Minimum score in the top 10% of this test that qualifies for the scholarship = ?
Answer:
Step-by-step explanation:
What is Normal Distribution?
We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.
We want to find out the minimum score that qualifies for the scholarship by scoring in the top 10% of this test.
The z-score corresponding to the probability of 0.90 is 1.28 (from the z-table)
Therefore, you need to score 1884 in order to qualify for the scholarship.
How to use z-table?
Step 1:
In the z-table, find the probability value of 0.90 and note down the value of the that row which is 1.2
Step 2:
Then look up at the top of z-table and note down the value of the that column which is 0.08
Step 3:
Finally, note down the intersection of step 1 and step 2 which is 1.28
Answer:
a) test statistic = 2.12
b) p-value = 0.017
c) we reject the Null hypothesis
Step-by-step explanation:
Given data :
N = 200
girls (x) = 115 , Boys = 85
p = x / n = 115 / 200 = 0.575
significance level ( ∝ ) = 0.1
<em>aim : test whether the proportion of girls births after the treatment is greater than 50% that occurs without any treatment </em>.
<u>A) Determine the test statistic </u>
H0 : p = 0.5
Ha : p > 0.5
to determine the test statistic we will apply the z distribution at ( ∝ ) = 0.1
Z - test statistic = ( 0.575 - 0.5) / = 2.12
<u>b) determine the p-value</u>
The P-value can be determined using the normal standard table
P-value = 1 - p(Z< 2.12 ) = 1 - 0.9830 = 0.017
c) Given that the p value ( 0.017 ) < significance level ( 0.1 )
we will reject the H0 because there is evidence showing that proportion of girls birth is > 50%