Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of
0 and a standard deviation of 1. Find the probability that a given score is less than 1.99 and draw a sketch of the region.
1 answer:
Answer:
Step-by-step explanation:
To find this probability, we shall be using the z-score route
Mathematically ;
z-score = (x -mean)/SD
From the question, x = 1.99, mean = 0 and SD = 1
So z = (1.99-0)/1 = 1.99
So the probability we want to calculate is;
P(z<1.99)
This value can be obtained from the standard normal distribution table.
P(z < 1.99) = 0.9767
The sketch of the region is as shown as in the attachment.
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<h3><u><em>Question:</em></u></h3>
On the first day, a total of 40 items were sold for $356. Pies cost $10 and cakes cost $8. Define the variables, write a system of equations to find the number of cakes and pies sold, and state how many pies were sold.
<h3><em><u>Answer:</u></em></h3>The variables are defined as:
"c" represent the number of cakes sold and "p" represent the number of pies sold
The system of equations used are:
c + p = 40 and 8c + 10p = 356
18 pies and 22 cakes were sold
<h3><em><u>Solution:</u></em></h3>Let "c" represent the number of cakes sold
Let "p" represent the number of pies sold
Cost of 1 pie = $ 10
Cost of 1 cake = $ 8
Given that total of 40 items were sold
number of cakes + number of pies = 40
c + p = 40 ------ eqn 1
<u><em>Given items were sold for $356</em></u>
number of cakes sold x Cost of 1 cake + number of pies sold x Cost of 1 cake = 356
8c + 10p = 356 ----- eqn 2
<u><em>Let us solve eqn 1 and eqn 2</em></u>
From eqn 1,
p = 40 - c ---- eqn 3
Substitute eqn 3 in eqn 2
8c + 10(40 - c) = 356
8c + 400 - 10c = 356
-2c = - 44
c = 22
<em>Substitute c = 22 in eqn 3</em>
p = 40 - c
p = 40 - 22
p = 18
Thus 18 pies and 22 cakes were sold