Answer:
(a) The area to the left of Z = -1.75 is 0.0401 or P(Z<-1.75) = 0.0401.
(b) The area to the left of Z = -0.31 is 0.3783 or P(Z<-0.31) = 0.3783.
(c) The area to the left of Z = -0.47 is 0.3192 or P(Z<-0.47) = 0.3192.
(d) The area to the left of Z = -0.56 is 0.2877 or P(Z<-0.56) = 0.2877.
Step-by-step explanation:
The question is asking the areas to the left of:
(a) Z = -1.75
(b) Z = -0.31
(c) Z = -0.47
(d) Z = -0.56
In other words, and mathematically:
[tex] \\ P(Z<-1.75)[/tex]
[tex] \\ P(Z<-0.31)[/tex]
[tex] \\ P(Z<-0.47)[/tex]
[tex] \\ P(Z<-0.56)[/tex]
To find these probabilities, we have to use the cumulative standard normal distribution. In this distribution, a special case for normal distributions in which the mean = 0 and the standard deviation = 1, the values are standardized, that is:
[tex] \\ Z = \frac{x - \mu}{\sigma}[/tex]
Where
x is a raw score (a value from a normally distributed data).
[tex] \\ \mu[/tex] is the population mean
[tex] \\ \sigma[/tex] is the population standard deviation
We can use the latter to "transform" any raw data coming from a normal distribution into z-scores. A z-score tells us the distance from the population mean in standard deviation units. A negative value for z means that it is located below the population mean. Conversely, a positive value for z indicates that it is located above the mean.
So, looking at all negative values for Z in the question, we can conclude that all values for Z are below the population mean.
Another important thing to be aware is that most cumulative standard normal distributions are made for positive values of Z. To overcome this, we know that all normal distributions are symmetrical around the mean, and so the standard normal distribution. In this case, to find probabilities for negative values of Z, we can use the next formula:
[tex] \\ P(Z<-a) = 1 - P(z<a) = P(z>a)[/tex] [1]
Having into account all the previous information, we can find all the probabilities using only the cumulative standard normal distribution (or, if wanted, a statistic program).
Then
(a) The area under the standard normal curve that lies to the left of Z = -1.75
[tex] \\ P(Z<-1.75)[/tex]
Consulting the cumulative standard normal distribution, and using [1], we have
[tex] \\ P(Z<-1.75) = 1 - P(Z<1.75)[/tex]
[tex] \\ P(Z<-1.75) = 1 - 0.95994[/tex]
[tex] \\ P(Z<-1.75) = 1 - 0.95994[/tex]
[tex] \\ P(Z<-1.75) = 0.04006[/tex]
Rounding to four decimal places:
The area to the left of Z = -1.75 is 0.0401 or P(Z<-1.75) = 0.0401.
(b) The area under the standard normal curve that lies to the left of Z = -0.31
[tex] \\ P(Z<-0.31)[/tex]
We can use the same procedure as the previous case. So
[tex] \\ P(Z<-0.31) = 1 - P(Z<0.31)[/tex]
[tex] \\ P(Z<-0.31) = 1 - 0.62172[/tex]
[tex] \\ P(Z<-0.31) = 0.37828[/tex]
Rounding to four decimal places:
The area to the left of Z = -0.31 is 0.3783 or P(Z<-0.31) = 0.3783.
(c) The area under the standard normal curve that lies to the left of Z = -0.47
[tex] \\ P(Z<-0.47)[/tex]
Following the same procedure as before:
[tex] \\ P(Z<-0.47) = 1 - P(Z<0.47)[/tex]
[tex] \\ P(Z<-0.47) = 1 - 0.68082[/tex]
[tex] \\ P(Z<-0.47) = 0.31918[/tex]
Rounding to four decimal places:
The area to the left of Z = -0.47 is 0.3192 or P(Z<-0.47) = 0.3192.
(d) The area under the standard normal curve that lies to the left of Z = -0.56
[tex] \\ P(Z<-0.56)[/tex]
We can use the same procedure as before:
[tex] \\ P(Z<-0.56) = 1 - P(Z<0.56)[/tex]
[tex] \\ P(Z<-0.56) = 1 - 0.71226[/tex]
[tex] \\ P(Z<-0.56) = 0.28774[/tex]
Rounding to four decimal places:
The area to the left of Z = -0.56 is 0.2877 or P(Z<-0.56) = 0.2877.
We can represent the four graphs corresponding to each case as shown below.
