Question

The weights of broilers (commercially raised chickens ) are approximately normally distributed with mean 1387 grams and standard deviation 161 grams . What is the probability that a randomly selected broiler weighs more than 1,532 grams ? Write only a number as your answer Round to 4 decimal places ( for example 0.0048 ). Do not write as a percentage .

The weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1387 grams and standard deviation 161 grams. What proportion of broilers weight between 1,453 and 1705 grams ? Write only a number as your answer. Round to 4 decimal places (for example 0.0048). Do not write as a percentage.

The lifetime of a certain type of automobile tire in thousands of miles) is normally distributed with mean 40 and standard deviation 7 Find the percentile corresponding to p=39% of the tire lifetimes. Write only a number as your answer. Round to two decimal places (for example: 0.81).

Electricity bills in a certain city have mean $ 92.55. Assume the bills are normally distributed with standard deviation $ 16.90. Find the value that separates the lower 72% of the bills from the rest.

A survey among freshmen at a certain university revealed that the number of hours spent studying the week before final exams was normally distributed with mean 38 and standard deviation 5 Find percentile corresponding to p=60% of the number of hours studying. Write only a number as your answer. Round to two decimal places (for example: 20.81 ).

Answer 1

Q1:

µ = 1387, σ = 161

P(X > 1532) =

= P( (X-µ)/σ > (1532-1387)/161)

= P(z > 0.9006)

= 1 - P(z < 0.9006)

Using excel function:

= 1 - NORM.S.DIST(0.9006, 1)

= 0.1839

Q2:

P(1453 < X < 1705) =

= P( (1453-1387)/161 < (X-µ)/σ < (1705-1387)/161 )

= P(0.4099 < z < 1.9752)

= P(z < 1.9752) - P(z < 0.4099)

Using excel function:

= NORM.S.DIST(1.9752, 1) - NORM.S.DIST(0.4099, 1)

= 0.3168

Q3:

µ = 40, σ = 7

Z score at p = 0.39 using excel = NORM.S.INV(0.39) = -0.2793

Value of X = µ + z*σ = 40 + (-0.2793)*7 = 38.04

Q4:

µ = 92.55, σ = 16.9

P(x < a) = 0.39

Z score at p = 0.39 using excel = NORM.S.INV(0.39) = -0.2793

Value of X = µ + z*σ = 92.55 + (-0.2793)*16.9 = 87.83

Q5:

µ = 38, σ = 5

Z score at p = 0.6 using excel = NORM.S.INV(0.6) = 0.2533

Value of X = µ + z*σ = 38 + (0.2533)*5 = 39.27