The amount of Jen's monthly phone bill is normally distributed with a mean of $59 and a standard deviation of $10. What percentage of her phone bills are between $29 and $89?

And we can find this probability with this difference:

And in order to find these probabilities using the normal standard distribution or excel and we got.  

So we expect about 99.735% of values between $29 and $89

Step-by-step explanation:

Let X the random variable that represent the amount of Jens monthly phone of a population, and for this case we know the distribution for X is given by:

Where and

We are interested on this probability first in order to find a %

The z score is given by:

If we apply this formula to our probability we got this:

And we can find this probability with this difference:

And in order to find these probabilities using the normal standard distribution or excel and we got.  

So we expect about 99.735% of values between $29 and $89

answer: a. 13/44

step-by-step explanation: judging by the information given to us, we are looking for experimental probability. we can see that 13 males have chosen chocolate ice cream. there are 44 ice creams in total (remember that the total is always the denominator). so our answer is 13/44.

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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