The average daily high temperature in Los Angeles is μ = 70°F with a standard deviation of σ= 5° F. Suppose that the temperatures in June closely follow a normal distribution: that is, if we let X be the temperature of a randomly chosen day in June, we have that X ~ N(μ, σ2) Required:
a. What is the probability of observing an 83℉ temperature or higher during a randomly chosen day in June?
b. What is instead the probability of observing a temperature between 70°F and 80°F?
c. How cold are the coldest 10% of the days during June in Los Angeles?
d. Let Y be temperature of a randomly chosen day in June measured in Celsius degrees (°C) instead of Fahrenheit (℉) degrees: that is, Y = (X-32) x . Write the probability model for Y. (In other words what is the probability distribution of Y?)
e. What is the probability of observing a 28.33°C (which corresponds to 83°F) temperature or higher in June in Los Angeles?

answer: i guess 105 degrees?

step-by-step explanation:

the largest number of the four options is 1.8 x 10-2

you have it in scientific notation.

1.80 x 10-6   = 0.00000180

1.80 x 10-4   = 0.000180 (bigger, or "less small", than the former one)

so

1.8 x 10-2       = 0.0180. as you see. the "less small", or the biggest one of them all.

as you have negative exponent, it means you are dividing 1.8 by 100 (i.e. moving the dot towards left, two times).

if you had positive exponent for the scientific notation, power of tenths, you would be multiplying times 100 (moving the dot towards right) as below:

1.8 x (10)2 = 180

so. if you have negative exponent in scientific notation, move the dot toward left. if it is positive, toward right as many times the exponent tells you.

answer:

step-by-step explanation:

here, the total dishes = 10

dishes having meat = 3 ( chicken, meatloaf and hamburger )

thus, the probability of having meat on the chosen dishes = number of dishes having meat / total dishes

=

therefore, the probability of not having meat = 1- the probability of having meat on the chosen dishes

=

=

=

⇒ option b is correct.