Question

The weight of a small starbucks coffee is a normally distributed random variable with a mean of 325 grams and a standard deviation of 10 grams. Find the weight that corresponds to each event (round your final answers to 2 decimal places)

Answer 1

The weight of a small starbucks coffee is a normally distributed random variable with a mean of 325 grams and a standard deviation of 10 grams. Find the weight that corresponds to each event (round your final answers to 2 decimal places)

a. Highest 10 percent

b. Middle 50 percent

Answer:

the weight that corresponds to Highest 10% = 337.8

the weight that corresponds to Middle 50 % lies between 318.26 and 331.74

Explanation:

From the information provided for us:

we have the mean = 325

the standard deviation = 10

The objective is to find the weight that corresponds to each event i.e for event (a) , highest 10%

So;

The probability of P (Z > z) = 10%

Same as:

0.1 = 1 - P( Z < z)

P( Z < z) = 1 - 0.1

P( Z < z) = 0.9

From the standard normal tables for z;

P( Z < 1.28) = 0.9

z = 1.28

Similarly. from the z formula; we have:

`z = (X - \mu)/(\sigma)`

`z * \sigma = X - \mu`

`z * \sigma + \mu= X`

`X= z * \sigma + \mu`

X = (1.28 × 10) + 325

X = 12.8 + 325

X = 337.8

Therefore, the weight that corresponds to Highest 10% = 337.8

b. the weight that corresponds to Middle 50 % can be computed as follows:

the region of z values at 0.50 lies between -0.674 and +0.674

from the z formula; we have:

`z = (X - \mu)/(\sigma)`

`z * \sigma = X - \mu`

`z * \sigma + \mu= X`

`X= z * \sigma + \mu`

X = -0.674 × 10 + 325 and X = 0.674 × 10 + 325

X = - 6.74 + 325 and X = 6.74 + 325

X = 318.26 and X = 331.74

the weight that corresponds to Middle 50 % lies between 318.26 and 331.74