6. Suppose that wedding costs in the Caribbean are normally distributed with a mean of $6000 and a standard deviation of $735. Estimate the percentage of Caribbean weddings that cost (a) between $5265 and $6735. % (b) above $6735. % (c) below $4530. % (d) between $5265 and $7470. %

6 Suppose that wedding costs in the Caribbean are normally distributed with a mean of 6000 and a standard deviation of 735 Estimate the percentage of Caribbean class=

Respuesta :

To solve this problem, the first thing we must do is find the Z-Score of the given costs: $5265 , $6735 , $4530 ,and $7470

Then we proceed to find the percentages for each interval based on the graph

z-score for $5265 )

[tex]Z_{5265}=\frac{5265-6000}{735}=-1[/tex]

z-score for $6735 )

[tex]Z_{6735}=\frac{6735-6000}{735}=1[/tex]

z-score for $4530 )

[tex]Z_{4530}=\frac{4530-6000}{735}=-2[/tex]

z-score $7470 )

[tex]Z_{7470}=\frac{7470-6000}{735}=2_{}[/tex]

now, let's analyze the intervals

a ) between $5265 and $6735

This interval goes from (μ-σ) to (μ+σ)

if we look at the graph we find that this corresponds to a percentage of 68%

b) above $6735

This corresponds to what is to the right of (μ+σ)

This is a percentage of 16%

[tex]\frac{100-68}{2}=\frac{32}{2}=16[/tex]

c ) below $4530

This corresponds to what is to the left of (μ-2σ)

This is a percentage of 2.5%

[tex]\frac{100-95}{2}=\frac{5}{2}=2.5[/tex]

d ) between $5265 and $7470

This interval goes from (μ-σ) to (μ+2σ)

This is a percentage of 81.5%

[tex]\begin{gathered} 100-\frac{100-68}{2}-\frac{100-95}{2} \\ =100-16-2.5 \\ =81.5 \end{gathered}[/tex]