Question

1. Suppose that the random variable X is normally distributed with mean μ = 30 and standard deviation σ = 4. Find a) P(x < 40) b) P(x > 21) c) P(30 < x < 35) 2. A radar unit is used to measure speeds of cars on a motorway. The speeds are normally distributed with a mean of 90 km/hr and a standard deviation of 10 km/hr. What is the probability that a car picked at random is travelling at more than 100 km/hr? 3. For a certain type of computers, the length of time between charges of the battery is normally distributed with a mean of 50 hours and a standard deviation of 15 hours. Find the probability that the length of time will be between 50 and 70 hours. 4. Entry to a certain University is determined by a national test. The scores on this test are normally distributed with a mean of 500 and a standard deviation of 100. Tom wants to be admitted to this university and he knows that he must score better than at least 70% of the students who took the test. Tom takes the test and scores 585. Will he be admitted to this university? 5. The annual salaries of employees in a large company are approximately normally distributed with a mean of $50,000 and a standard deviation of $20,000. a) What percent of people earn less than $40,000? b) What percent of people earn between $45,000 and $65,000? c) What percent of people earn more than $70,000?

Answer 1

Question 1) - X follows Normal(=30,=4)

According to the properties of Normal distribution Z = follows N(0,1) distribution.

a) P[ X<40 ] = = = 0.9938

b) P[ X>21 ] = = P [ z > -2.25] = P[ z< 2.25] = 0.9878

#due to symmetry of normal distribution

c) P[ 30<X<35 ] = = P[ 0 < Z < 1.25] = P[ Z<1.25] - P[Z<0] = 0.8944 - 0.5

= 0.3944

Question 2)-  Speeds are normally distributed with =90 km/hr and =10 km/hr

According to the properties of Normal distribution Z = follows N(0,1) distribution.

Probability that car picked at random is travelling at more than 100 km/hr = P[ X>100]

= = P[ Z > 1] = 1 - P[Z<1] = 1 - 0.8413 = 0.1587

Question 3) - Let X be length of time between charges of the battery.

X follows N( =50 hrs ,=15 hrs)

According to the properties of Normal distribution Z = follows N(0,1) distribution.

Probability that length of time is between 50 and 70 hrs = P[ 50<X<70 ] =

= P[ 0<Z<1.33 ] = P[ Z<1.33] - P[Z<0] = 0.9082 - 0.5 = 0.4082

Question 4)- Let X denote scores of the national test

X follows N( =500 marks ,=100 marks)

According to the properties of Normal distribution Z = follows N(0,1) distribution.

We need to find the point "a" such that 70% of the students score less than that marks.

P[ X<a ] = 0.7

= 0.7

P[ Z < ] = 0.7

= 0.52

a = 100*0.52 + 500 = 552

Hence, a student requires at least 552 marks to score better than 70% of the students who appeared for the test.

Tom scored 585. That is Tom scored better than 70% of the students who appeared for the test. Hence, Tom will be admitted to the university.

Question 5) - Let X denote the annual salary of the employees.

X follows N( = $50,000 ,=$20,000 )

According to the properties of Normal distribution Z = follows N(0,1) distribution.

a)  P[ X<40,000] = = P[ Z< -0.5] = 1 - P[ Z> -0.5] = 1 - P[ Z<0.5]

= 1 - 0.6915 = 0.3085

30.85% of people earn less than $40,000.

b) P[ 45,000<X<65,000 ] = = P[ -0.25< Z < 0.75] = P[ Z<0.75] - P[ Z< -0.25] = P[ Z<0.75] - (1 - P[ Z< 0.25]) = 0.7734 - ( 1- 0.5987) = 0.3721

37.21% of people earn between $45,000 and $65,000.

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