Question

The weight of a small Starbucks coffee is a normally distributed random variable with a mean of 415 grams and a standard deviation of 23 grams. Find the weight that corresponds to each event.

Answer 1

Answer:

(a) The highest 20% weight correspond to the weight 434.32 grams.

(b) The middle 60% weight correspond to the weights 434.32 grams and 395.68 grams.

(c) The highest 80% weight correspond to the weight 395.68 grams.

(d) The highest 80% weight correspond to the weight 391.08 grams.

Step-by-step explanation:

Let X = weight of a small Starbucks coffee.

It is provided: $X\sim N(\mu = 415\ grams, \sigma=23\ grams)$

(a)

Compute the value of x fro P (X > x) = 0.20 as follows:

$P (X>x)=0.20\\P(\frac{X-\mu}{\sigma}>\frac{x-415}{23} )=0.20\\P (Z>z)=0.20\\1-P(Z<z)=0.20\\P(Z<z)=0.80$

Use a standard normal table.

The value of z is 0.84.

The value of x is:

$0.84=\frac{x-415}{23}\\0.84\times23=x-415\\x=415+19.32\\=434.32$

Thus, the highest 20% weight correspond to the weight 434.32 grams.

(b)

Compute the value of x fro P (x₁ < X < x₂) = 0.60 as follows:

$P(x_{1}<X<x_{2})=0.60\\P(\frac{x_{1}-415}{23}<\frac{X-\mu}{\sigma}< \frac{x_{2}-415}{23})=0.60\\P(-z<Z<z)=0.60\\P(Z<z)-P(Z<-z)=0.60\\P(Z<z)-[1-P(Z<z)]=0.60\\2P(Z<z)=1.60\\P(Z<z)=0.80$

Use a standard normal table.

The value of z is 0.84.

The value of x₁ and x₂ are:

$z=\frac{x_{1}-415}{23}\\0.84=\frac{x_{1}-415}{23}\\x_{1}=415+(0.84\times23)\\=434.32$

$-z=\frac{x_{2}-415}{23}\\0.84=\frac{x_{2}-415}{23}\\x_{1}=415-(0.84\times23)\\=395.68$

Thus, the middle 60% weight correspond to the weights 434.32 grams and 395.68 grams.

(c)

Compute the value of x fro P (X > x) = 0.80 as follows:

$P (X>x)=0.80\\P(\frac{X-\mu}{\sigma}>\frac{x-415}{23} )=0.80\\P (Z>z)=0.80\\1-P(Z<z)=0.80\\P(Z<z)=0.20$

Use a standard normal table.

The value of z is -0.84.

The value of x is:

$-0.84=\frac{x-415}{23}\\-0.84\times23=x-415\\x=415-19.32\\=395.68$

Thus, the highest 80% weight correspond to the weight 395.68 grams.

(d)

Compute the value of x fro P (X < x) = 0.15 as follows:

$P (X<x)=0.15\\P(\frac{X-\mu}{\sigma}<\frac{x-415}{23} )=0.15\\P (Z<z)=0.15$

Use a standard normal table.

The value of z is -1.04.

The value of x is:

$-1.04=\frac{x-415}{23}\\-1.04\times23=x-415\\x=415-23.92\\=391.08$

Thus, the highest 80% weight correspond to the weight 391.08 grams.