Given a normal distribution with 49 and 55 complete parts a

Given a normal distribution with

=49

and

=55,

complete parts (a) through (d)

a.

What is the probability that

X>42?

P(X>42)=

(Round to four decimal places as needed.)

b.

What is the probability that

X<39?

P(X<39)=

(Round to four decimal places as needed.)

c.

For this distribution,

99% of the values are less than what X-value?

X=

(Round to the nearest integer as needed.)

d.

Between what two X-values (symmetrically distributed around the mean) are

80%of the values? For this distribution,80% of the values are between X=nothing

and X=

(Round to the nearest integer as needed.)

Solution

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    42      
u = mean =    49      
          
s = standard deviation =    55      
          
Thus,          
          
z = (x - u) / s =    -0.127272727      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -0.127272727   ) =    0.550637727 [ANSWER]

*********************

b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    39      
u = mean =    49      
          
s = standard deviation =    55      
          
Thus,          
          
z = (x - u) / s =    -0.181818182      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.181818182   ) =    0.427862708 [ANSWER]

*******************

c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.99      
          
Then, using table or technology,          
          
z =    2.326347874      
          
As x = u + z * s,          
          
where          
          
u = mean =    49      
z = the critical z score =    2.326347874      
s = standard deviation =    55      
          
Then          
          
x = critical value =    176.9491331   [ANSWER]

************************

d)

80% of the middle values are bounded by the 10th and 90th percentile.  

For the 10th percentile:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.1      
          
Then, using table or technology,          
          
z =    -1.281551566      
          
As x = u + z * s,          
          
where          
          
u = mean =    49      
z = the critical z score =    -1.281551566      
s = standard deviation =    55      
          
Then          
          
x = critical value =    -21.4853361      

For the 90th percentile:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.9      
          
Then, using table or technology,          
          
z =    1.281551566      
          
As x = u + z * s,          
          
where          
          
u = mean =    49      
z = the critical z score =    1.281551566      
s = standard deviation =    55      
          
Then          
          
x = critical value =    119.4853361      

Thus, they are between (rounding) -21 and 119. [ANSWER]