Question

27. The amount of paste in a tube of special crème dispensed by a machine follows distribution. The Mean of the distribution is 150mg and the Standard Deviation is 5mg. that the machine will dispense between 150mg and 154mg? a normal probability What is the probability a. 0.2881 b. 0.5 c. 1 d. 0.1550 e. 0.0462 distribution. The Mean of the distribution is 150 Fahrenheit and the Standard Deviation is 5 Fahrenheit. What is the probability that the temperature of the Coffee Macchiato is more than 164 Fahrenheit? 28. The lev el of temperature in a Coffee Macchiato at the Choy Coffee Shop follows a normal probability a. 0.0947 b. 0.4974 c. 0.0026 d. 0.5 e. 0.2881 You randomly test 10

Answer 1

(A)

Here we have given that

$\mu$ = population mean =150

$\sigma$ = population standard deviation=5

Now we want to find

The Probablity that the machine will dispense between 150 and 153 mg

i.e P(150 < X < 154)

For that 1^st we want to find the Zscore

For X=150

$Zscore = \frac {x -\mu} { \sigma}$

= $\frac {150 -150} {5}$

=0.00

Now, For x= 154

Zscore = $\frac {x -\mu} { \sigma}$

= $\frac {154 -150} {5}$

             =-0.80

i.e we get

                                                      =0.2119 – 0.5000 ( using z standard normal talble)

                                                       =0.2881

That is

We get

P(150 < X < 154) =0.2881

that is here option A is correct

(B)

Here we have given that

$\mu$ = population mean =150

$\sigma$ = population standard deviation=5

Now we want to find

The Probablity that the temperature of the coffee macchiato is more than 164 Fahrenheit.

i.e P( X> 164)

For that 1^st we want to find the Zscore

For X=164

$Zscore = \frac {x -\mu} { \sigma}$

= $\frac {164 -150} {5}$

=2.80

i.e we get

P( Z > 2.80) =1-P( Z < 2.80)

                     =1-0.9974 ( using standard normal Z table)

                      =0.0026

i.e we get

P(Z > 164) = 0.0026

That is here option C is correct.