Answer:
Step-by-step explanation:
0.178
Answer:
8304
Step-by-step explanation:
1 box=48 books
173 boxes=48x173
=8304
Using the normal distribution, it is found that:
- 3 - a) The 40th percentile of the height of Dinaric Alps distribution for men is of 72.2 inches.
- 3 - b) The minimum height of man in the Dinaric Alps that would place him in the top 10% of all heights is of 76.84 inches.
- 4 - a) The 25th percentile for the math scores was of 71.6 inches.
- 4 - b) The 75th percentile for the math scores was of 78.4 inches.
<h3>Normal Probability Distribution
</h3>
In a <em>normal distribution </em>with mean
and standard deviation
, the z-score of a measure X is given by:
- 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.
Question 3:
- The mean is of 73 inches, hence
.
- The standard deviation is of 3 inches, hence
.
Item a:
The 40th percentile is X when Z has a p-value of 0.4, so <u>X when Z = -0.253</u>.




The 40th percentile of the height of Dinaric Alps distribution for men is of 72.2 inches.
Item b:
The minimum height is the 100 - 10 = 90th percentile is X when Z has a p-value of 0.9, so <u>X when Z = 1.28</u>.




The minimum height of man in the Dinaric Alps that would place him in the top 10% of all heights is of 76.84 inches.
Question 4:
- The mean score is of 75, hence
.
- The standard deviation is of 5, hence
.
Item a:
The 25th percentile is X when Z has a p-value of 0.25, so <u>X when Z = -0.675</u>.




The 25th percentile for the math scores was of 71.6 inches.
Item b:
The 75th percentile is X when Z has a p-value of 0.25, so <u>X when Z = 0.675</u>.




The 75th percentile for the math scores was of 78.4 inches.
To learn more about the normal distribution, you can take a look at brainly.com/question/24663213
Y=-2x+0 reason: rise over run as you can see in the graph -2 rise (drop) and -1 run and b=0 y intercept
Step-by-step explanation:
S = { 1, 2, 3, 4, 5, 6 7, 8 }
n ( S ) = 8
Let A be the event of getting 4,
A = { 4 }
n ( A ) = 1
P ( A )
= n ( A ) / n ( S )
= 1 / 8
Therefore, the probability of spinning a 4 is 1 / 8.
S = { A, B, A, C, A, B }
n ( S ) = 6
Let Y be the event of getting C,
Y = { C }
n ( Y ) = 1
P ( Y )
= n ( Y ) / n ( S )
= 1 / 6
Therefore, the probability of spinning a C is 1 / 6.