1. During the past 6 months, 73.2% of US households purchased sugar. Assume that

Question

1. During the past 6 months, 73.2% of US households purchased sugar. Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10. 99% of the households spent less than what amount? Please show your work in detail.

mean = $8.22

standard deviation = $1.10

z = x – mean / standard deviation

2.33 = x – $8.22 / $1.10

2.33 x $1.10 = x - $8.22

2.563 = x - $8.22

2.563 + $8.22 = x

10.783 = x

I know the answer is correct. I just need help with figuring out z = 2.33. How do you find that out.

2. During the past 6 months, 73.2% of US households purchased sugar. Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10. What proportion of the households spent between $5.00 and $9.00 on sugar? Please show your work in detail.

mean, m = $8.22

standard deviation, s = $1.10

proportion of the households spent between $5 and $9

P (5 < x < 9)

P = ($5 – $8.22 / $1.10 < x - m / s < $9 - $8.22 / $1.10)

P = (-2.93 < z < 0.71)

P = (z < 0.71) – P(z < -2.93)

P = 0.7611 – 0.0017

P = 0.7594

For the second to last step, how do you do the math to find P = 0.7611 - 0.0017. Please show me the steps to finding 0.7611 and 0.0017

Explanation / Answer

1:

Here you need z-score that has 0.99 area to its left. In the z-table, go to row 2.3 and column 0.03 you get 0.9901. That is z-score 2.33 has 0.9901 area to its left.

2:

P(z < 0.71) shows the area left to z-score 0.71. In the z table, go to row 0.7 and column 0.01 you will get 0.7611.

P(z < -2.93) shows the area left to z-score -2.93. In the z table, go to row -2.9 and column 0.03 you will get 0.0017.

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