Question

At a large bank, account balances are normally distributed with a mean of $1,637.52 and a standard deviation of $623.16. What is the probability that a simple random sample of 400 accounts has a mean that exceeds $1,650?

Answer 1

Answer: the probability is 0.49

Step-by-step explanation:

Since the account balances at the large bank are normally distributed.

we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = account balances.

µ = mean account balance.

σ = standard deviation

From the information given,

µ = $1,637.52

σ = $623.16

We want to find the probability that a simple random sample of 400 accounts has a mean that exceeds $1,650. It is expressed as

P(x > 1650) = 1 - P(x ≤ 1650)

For x = 1650,

z = (1650 - 1637.52)/623.16 = 0.02

Looking at the normal distribution table, the probability corresponding to the z score is 0.51

P(x > 1650) = 1 - 0.51 = 0.49

Answer 2

Answer:

$P(\bar X >1650)=P(Z>\frac{1650-1637.52}{\frac{623.16}{\sqrt{400}}}=0.401)$

And we can use the complement rule and we got:

$P(Z>0.401) =1-P(Z$

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the bank account balances of a population, and for this case we know the distribution for X is given by:

$X \sim N(1637.52,623.16)$  

Where $\mu=1637.52$ and $\sigma=623.16$

Since the distribution of X is normal then the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using this formula we got:

$P(\bar X >1650)=P(Z>\frac{1650-1637.52}{\frac{623.16}{\sqrt{400}}}=0.401)$

And we can use the complement rule and we got:

$P(Z>0.401) =1-P(Z$