Question

2. If the difference between the sample mean and the population mean is a result of recording the results of a survey incorrectly, then this is referred to as what type of error.

b) Find the area under the standard normal curve from z = 0 to z = 4.35

c) Find the value of z so that the area under the standard normal curve in the right tail is .0351

d) determine the area under a normal distribution curve to the left of x = 22 if the population mean is 55 and population Standard deviation is 7.

Answer 1

2.

a. Solution – Sampling Error

Explanation:

Sampling error occurs only as a result of using a sample from a population, rather than conducting a census (complete enumeration) of the population. It refers to the difference between an estimate for a population based on data from a sample and the 'true' value for that population.

Hence the difference between the sample mean and the population mean as a result of recording the results of a survey incorrectly is called as sampling error

b.Solution is 0.4999

Explanation:

Required probability is given by P(0<Z<4.35)

Required probability is shown in the below normal curve

Using the Z score calculator, we get the required probability = 0.4999

c.Solution is 1.81

Area under the standard normal curve in the right tail is .0351. The same is represented in the below normal curve:

Based on the above data we can tell that the right side remaining area will be 1-0.0351 = 0.9649

Now we know that standard curve has total area of 1 and it is symmetrical about mean (mean =0).

So 0.9649-0.5 = 0.4649

Now required Z value is nothing that calculating Z1 value such that the area between Z= 0 and Z=Z1 [i.e. P(0<Z<Z1) = 0.4649] is equal to 0.4649

Using the Z score calculator/graph we have Z1 value given by 1.81

d.Solution is close to 0

Population Mean (μ) = 55

Population Standard deviation (σ) = 7

Also X = 22

Formula to calculate Z statistic is given by (X- μ)/ σ

= (22-55)/7

=-33/7

Z=-4.71

So Required area is P(Z<-4.71) = P(Z>4.71)

This value is close to 0