How to find the area under the standard normal curve between z 1.5 and z 2.5?

1 answer:

8 0

It is possible to calculate mathematically the area under the normal curve between any two values of z.
However, tables/software have been developed to give the areas under the normal curve to the left of particular values of z. The function is the probability of Z<z, or P(Z<z).
The area between two values z1 and z2 (where z2>z1) is therefore
P(Z<z2)-P(Z<z1).

For example, to find the area between z1=1.5, z2=2.5
is
P(Z<2.5)-P(Z<1.5)
=0.99379-0.93319
=0.06060

(above values obtained by software, such as R)
For example, the value P(Z<2.5) can be calculated using
P(Z<2.5)=erf(2.5/sqrt(2))/2+1/2
where erf(x) is a mathematical function that does not have an explicit formula (calculated by summation of series, or tabulated).

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a) $\hat p=\frac{471}{1024}=0.460$

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$SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156$

And the margin of error is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0305$

b) The 99% confidence interval would be given by (0.429;0.491)

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Data given and notation

n=1024 represent the random sample taken

X=471 represent the people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

$\hat p=\frac{471}{1024}=0.460$ estimated proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

p= population proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

Part a

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.