Assume there is a mound shaped distribution with a mean of 56 and a standard dev
ID: 3272222 • Letter: A
Question
Assume there is a mound shaped distribution with a mean of 56 and a standard deviation of 2.5. Use this information to answer the following questions. What is the approximate percent of the area between 51 and 61? What is the approximate percent of the area below 53.5? What is the approximate percent of the area between 61? What is the approximate percent of the area between 58.5| and 63.5? What is the z-score for 63.3? What is the z-score for 61? What is the data values associated with the z-score of -1.5? What is the data value associated with the z-score of 2.2?Explanation / Answer
Solution3c:
mean=56
sd=2.5
P(51<X<61)
z=x-mean/sd
for x=51
z=51-56/2.5
z=-2
for x=61
z=x-mean/sd
=61-56/2.5
=5/2.5
=2
s0
P(-2<z<2) to be found
Step 2: To find the probability of P (2<Z<2), we use the following formula:
P (2<Z<2 )=P ( Z<2 )P (Z<2 )
Step 3: P ( Z<2 ) can be found by using the following standard normal table.
From Standard Normal Table
We see that P ( Z<2 )=0.9772.
Step 4: P ( Z<2 ) can be found by using the following fomula.
P ( Z<a)=1P ( Z<a )
After substituting a=2 we have:
P ( Z<2)=1P ( Z<2 )
P ( Z<2 ) can be found by using the following standard normal table.
FromStandard Normal Table
We see that P ( Z<2 )=0.9772 so,
P ( Z<2)=1P ( Z<2 )=10.9772=0.0228
At the end we have:
P (2<Z<2 )=0.9544
:
so answer is 0.9544*100=95.44%
Solution3d:
P(X<53.5)
z=x-mean/sd
z=53.5-56/2.5
=-2.5/2.5
=-1
P(Z<-1) to be found
P ( Z<1 )=1P ( Z<1 )=10.8413=0.1587
ANSWER IS 0.1587*100=15.87%
Solution3e:
P(X>61)
z=x-mean/sd
z=61-56/2.5
=5/2.5
=2
P(Z>2 ) to be found
P ( Z>2 )=1P ( Z<2 )=10.9772=0.0228
so ANSWER IS0.0228*100=2.28%
Solution3f:
P(58.5<x<63.5)
=P(58.5-56/2.5<z<63.5-56/2.5)
=P(1<z<3)
Step 2: To find the probability of P (1<Z<3), we use the following formula:
P (1<Z<3 )=P ( Z<3 )P (Z<1 )
Step 3: P ( Z<3 ) can be found by using the following standard normal table.
From Standard Normal Table
We see that P ( Z<3 )=0.9987.
Step 4: P ( Z<1 ) can be found by using the following standard normal table.
From Standard Normal Table
We see that P ( Z<1 )=0.8413.
At the end we have:
P (1<Z<3 )=0.1574
SO ANSWER IS
0.1574*100=15.74%