Question

1)Dog weights. Adult German shepherd weights are normally distributed with mean of 73 pounds and standard deviation of 8 pounds.

(a) The bottom 24% of weights are below what weight? _________

(b) 76% of weights are above what weight?___________

(c) The top 24% of weights are above what weight? ___________

(Round answers to one decimal place)

2)A distribution of values is normal with a mean of 60 and a standard deviation of 7.

Find the interval containing the middle-most 82% of scores: _________

Enter your answer accurate to 1 decimal place using interval notation. Example: (2.1,5.6)

3)A particular fruit's weights are normally distributed, with a mean of 274 grams and a standard deviation of 35 grams.

The heaviest 6% of fruits weigh more than how many grams?

Give your answer to the nearest gram.

____________

4)A company produces packets of soap powder labeled "Giant Size 20 oz." The actual weight of soap powder in such a box has a Normal distribution with a mean of 21 oz and a standard deviation of 0.7 oz. To avoid dissatisfied customers, a box of soap is considered underweight if it weighs less than 20 oz. To avoid losing money, the top 5% (the heaviest 5%) is labeled overweight.

How heavy does a box have to be in order for the box to be labeled overweight?

______________oz Round to 2 places.

Answer 1

Answer:

1.

Given,

Mean = 73

Standard deviation = 8

P(Z < z) = 0.24

since from standard normal table

z = - 0.71

(x - u)/s = - 0.71

(x - 73)/8 = - 0.71

x = 67.32

b)

P(Z > z) = 0.76

1 - P(Z < z) = 0.78

P(Z < z) = 0.22

since from standard normal table

z = - 0.77

(X - 73)/8 = - 0.77

x = 66.84

c)

P(Z > z) = 0.24

since from standard normal table

z = 0.71

(x - 73)/8 = 0.71

x = 78.68

2.

Mean = 60

Standard deviation = 7

P(-z < Z < z) = 0.82

2P(Z < z) = 1 + 0.82 = 1.82

P(Z < z) = 1.82/2 = 0.91

since from standard normal table

P(Z < 0.55) = 0.91

z = +/- 0.55

(x - 60)/7 = 0.55

x = 63.85

(x - 60)/7 = - 0.55

x = 56.15

Middle most 82% are from 56.15 to 63.85

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