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A survey by the National Retail Federation found that women spend on average $146.21 for the Christmas holidays. Assume that standard deviation is $29.44. Find the percentage of women who spend less than $160.00. Assume the variable is normally distributed.A survey by the National Retail Federation found that women spend on average $146.21 for the Christmas holidays. Assume that standard deviation is $29.44. Find the percentage of women who spend less than $160.00. Assume the variable is normally distributed.

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JeanaShupp

Answer: 68.03%

Step-by-step explanation:

Given, A survey by the National Retail Federation found that women spend on average ([tex]\mu[/tex]) $146.21 for the Christmas holidays.

[tex]\sigma=\$29.44[/tex]

Let x be a random variable that denotes the amount that woman spend for the Christmas holidays. .

We assume that x is normally distributed.

Now , the percentage of women who spend less than $160.00 would be:

[tex]P(x<\$160)=P(\dfrac{x-\mu}{\sigma}}<\dfrac{160-146.21}{29.44})[/tex]

[tex]=P(z<0.468)[\because \ z=\dfrac{x-\mu}{\sigma}]\\\\=0.6803\approx68.03\%[/tex][By z-table]

Hence, the percentage of women who spend less than $160.00.  = 68.03%

The percentage of women who spend less than $160.00. Assume the variable is normally distributed is 68.03%

  • The calculation is as follows:

[tex]P(x<\$160) = P(\frac{x-\mu}{\sigma} <\frac{160-146.21}{29.44})\\\\= P(z>0.468)[/tex]

= 0.6803

= 68.03%

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