Question

Suppose the mean starting salary for nurses is $67,709 nationally. The standard deviation is approximately $10,970. Assume that the starting salary is normally distributed.

Find the probability that a starting nurse will make more than $88,000. Round to four decimals.
P(x > $88,000) =

Find the probability that a starting nurse will make less than $58,000. Round to four decimals.
P(x < $58,000) =

Find the probability that a starting nurse will make between $67,000 and $70,000. Round to four decimals.
P($67,000 < x < $70,000) =

What salary do 30% of all nurses make more than? Round to the nearest dollar.
$______

B) Suppose the mean yearly rainfall for a city is about 130 mm and the standard deviation is about 72 mm. Assume rainfall is normally distributed.

Find the probability that the yearly rainfall is less than 93 mm. Round to four decimal places.
P(x < 93) =

Find the probability that the yearly rainfall is more than 213 mm.
P(x > 213) =

Find the probability that the yearly rainfall is between 188 and 238 mm.
P(188 < x < 238) =

What rainfall amount are 90% of all yearly rainfalls more than? Round to the nearest whole number.

Answer 1

Solution:
Given that,
μ =$67,709 , σ=$10,970
a) P(X>88000)=1-P(X<=88000)
=1- p{[(x- μ)/σ]<=[(88000 - 67709)/10970]}
= 1- P(z<=1.85)
=1- 0.9678( from Standard Normal table)
=0.0322
P(X>88000)=0.0322
b) P(X<58000)

= p{[(x- μ)/σ]<[(58000 - 67709)/10970]}
=P(z< -0.88)
P(X< 58000)=0.1894 ( from Standard Normal table)
C)
P(67000<X< 70000)

= p{[(67000- 67709)/10970]<[(X- μ)/σ]<[(70000 - 67709)/10970]}
=P(-0.06<Z< 0.21)
= p(Z< 0.21) - p(Z< -0.06)
= 0.5832 - 0.4761 ( from. Standard Normal table)
=0.1071
P(67000<X< 70000)= 0.1071
d)Let' a 'be the salary do 30% of all nurses make more than.
P( X> a)= 0.30
1-P( X<=a)= 0.30
1-p{[(X- μ)/σ]<=[(a - 67709)/10970]}=0.30
1-P( Z<=z)= 0.30 .... where z=(a - 67709)/10970
P( Z<=z)= 0.70
By using the standard normal table , we get
P( Z<=0.52)= 0.70
z= 0.52
Now using the z-score formula, we can find the value of ' a ' as follows,
z=(a- μ)/σ
z=(a - 67709)/10970
(0.52)×10970=( a - 67709)
a= 67709+[(0.52)×10970]
=67709+5704.4
a= 73413.4
a=$ 73413 .... nearest dollar.
P( X< $73413)= 0.30

B)Solution:
Given that,
μ =130 mm, σ= 72 mm
a) P(X<93)= p{[(x- μ)/σ]<[(93 - 130)/72]}
=P(z< -0.51)
P(X< 93)=0.3050 ( from Standard Normal table)
b) P(X>213)=1-P(X<=213)
=1- p{[(x- μ)/σ]<=[(213 - 130)/72]}
= 1- P(z<=1.15)
=1- 0.8749( from Standard Normal table)
=0.1251
P(X>213)=0.1251

C)
P(188<X< 238)= p{[(188- 130)/72]<[(X- μ)/σ]<[(238 - 130)/72]}
=P(0.80<Z< 1.5)
= p(Z< 1.5) - p(Z< 0.80)
= 0.9332 - 0.7881 ( from. Standard Normal table)
=0.1451
P(188<X< 238)= 0.1451
d)Let' a 'be the rainfall amount are 90% of all yearly rainfalls more than.
P( X> a)= 0.90
1-P( X<=a)= 0.90
1-p{[(X- μ)/σ]<=[(a - 130)/72]}=0.90
1-P( Z<=z)= 0.90 .... where z=(a - 130)/72
P( Z<=z)= 0.10
By using the standard normal table , we get
P( Z<=-1.28)= 0.10
z= -1.28
Now using the z-score formula, we can find the value of ' a ' as follows,
z=(a- μ)/σ
z=(a - 130)/72
(-1.28)×72=( a - 130)
a= 130+[(-1.28)×72]
=130-92.16
a= 37.84
a=38 .... nearest whole number.
P( X< 38)= 0.90