Question

Part 1)

A recent study found that we now spend more time looking at our phone/tablet screens than at a TV screen. The amount of time spent looking at phone/tablet screens is normally distributed with an average of 180 minutes a day with a standard deviation of 20 minutes. What is the lowest and highest amount of time spent on the phone/tablet screens for the middle 60%?

Johnny scored a 115 on an IQ test that was normally distributed with a mean of 100 and standard deviation of 20. What would his equivalent score be on a different IQ test that has a mean 85 and standard deviation of 12?

Part 2)

A college math placement test has a mean of 250 and a standard deviation of 15. If the top 16% are exempt from the math requirement, what score does an incoming freshmen need so that they do not have to take a math class?

Companies who design furniture for elementary classrooms produce a variety of sizes for kids of different ages. Suppose the heights of kindergarten children are normally distributed with a mean of 38.2 inches. If it is known that 40.3 inches represents the largest 10% of kindergarten children’s heights, what is the standard deviation?

Please answer all parts I'd really appreciate it. Thank you.

Answer 1

Part 1)

a) Let x be the amount of time spent looking at phone/tablet screens.

mean ( µ ) = 180 and standard deviation (σ) = 20

We are asked to find x1 and x2 such that are between them is 60% or 0.60

So area outside the x1 and x2 is 1 - 0.60 = 0.40 , this area equally divided into two parts at both tails ,

So area under lower tail and above upper tail is 0.40/2 = 0.20

Z score for approximate area 0.2000 on the -0.84 on z score table

x1 = (z*σ ) + µ = ( -0.84*20 ) + 180

x1 = 163.2

x1 = (z*σ ) + µ = ( 0.84*20 ) + 180

x1 = 196.8

The lowest and highest amount of time spent on the phone/tablet screens for the middle 60% is 163.2 and 196.8

b)

Let x be the IQ score on 1st test and y be the IQ score on the 2nd test

= 100 , = 20 and = 85 , = 12

z =

z = 0.75

y = (z*) + = ( 0.75*12 ) + 85

y = 94

So his equivalent score on a different IQ test would be 94

Part 2a)

Let x be the math score on test

mean ( µ ) = 250 and standard deviation (σ) = 15

We have to find x such that area above x is 0.16, or area below x is 1 - 0.16 = 0.84

So z score corresponding to approximate area 0.8400 is 0.99 on z score table

x = (z*σ ) + µ = ( 0.99*15) + 250

x = 264.85

So freshmen need score 264.85 so that they do not have to take a math class

b)

Let x be the heights of kindergarten children

mean ( µ ) = 38.2  

We are given that area above x = 40.3 is 0.10, so area below 40.3 is 1-0.10 = 0.90

Therefore z score corresponding to approximate area 0.90 is 1.28 on z score table

σ = =

σ = 1.64

The standard deviation is 1.64