Question

Suppose that the probability of getting an interview when applying for a professional position is 0.15. You apply for 40 jobs. Use the normal distribution to estimate the probability that more than 5 of the applications result in interviews. Round to four decimal places.

Answer 1

Answer:

0.6700 = 67% probability that more than 5 of the applications result in interviews.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$n = 40, p = 0.15$

So

$\mu = E(X) = np = 40*0.15 = 6$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{40*0.15*0.75} = 2.26$

Probability that more than 5 of the applications result in interviews.

This is 1 subtracted by the pvalue of Z when X = 5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5 - 6}{2.26}$

$Z = -0.44$

$Z = -0.44$ has a pvalue of 0.33

1 - 0.33 = 0.6700

0.6700 = 67% probability that more than 5 of the applications result in interviews.

Answer 2

The probability that more than 5 of the applications result in interviews will be "0.5876". To understand the calculation, check below.

Probability

According to the question,

Number of jobs, n = 40

Interview getting probability, p = 0.15

Now,

Mean = np

          = 40 × 0.15

          = 6

Standard deviation be:

= √40 × 0.15 (1 - 0.15)

= 2.258

The z-value be:

= $\frac{5.5-6}{2.258}$

= -0.211

hence,

The probability will be:

→ P (z > -0.211) = 0.5876

Thus the above answer is correct.

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