Question

Four friends, Janice, Barbara, Kathy and Roberta, decided to carpool together to get to school. Each day the driver would be chosen by randomly selecting one of the four names. They carpool to school for $96$days. Use the normal approximation to the binomial to calculate the following probabilities. Round the standard deviation to four decimal places.

a. Find the probability that Janice is the driver at most$20$days.

b. Find the probability that Roberta is the driver more than $16$days.

c. Find the probability that Barbara drives exactly $24$ of those $96$ days.

Short answer

(a) The probability that Janice is the driver at most 20 days is$0.2033$

(b) The probability that Roberta is the driver more than 16 days is$0.9772$

(c) The probability that Barbara drives exactly 24 of those 96 days is$0.0956$

Step-by-step solution

Part (a) Step1: Given information 

Given in the question that, Four friends, Janice, Barbara, Kathy and Roberta, decided to carpool together to get to school. Each day the driver would be chosen by randomly selecting one of the four names. They carpool to school for$96$days.

Part (a) Step 2: Explanation

From the information, we observed that, the sample size is $96$days.

Therefore,

$X~B(96,0.25)$

Since,$np=24>5$and$nq=72>5$

Hence,

The mean:

$\mu =n\cdot p$

$=96\cdot 0.25$

$=24$

Standard Deviation

$s=\sqrt{npq}=4.24$

Now random variable $Y$has a normal distribution

$Y~N(24,4.24)$

The wanted probability is given by:

$P(X<20)=P(Y<20.5)$

$=P\left(\frac{Y-24}{4.24}<\frac{20.5-24}{4.24}\right)$

$=\varphi \left(\frac{20.5-24}{4.24}\right)$

$=\varphi (-0.83)$

$=1-0.7967$

$=0.2033$

Part (b) Step 1: Given information

Given in the question that, Four friends, Janice, Barbara, Kathy and Roberta, decided to carpool together to get to school. Each day the driver would be chosen by randomly selecting one of the four names. They carpool to school for $96$days.

Part (b) Step 2: Explanation

Variable $X$has a binomial distribution:

$X~B(96,0.25)$

The mean:

localid="1648366098955" $\mu =n·p=96·0.25=24$

Standard deviation:

$s=\sqrt{npq}=4.24$

Now random variable $Y$has a normal distribution:

$Y~N(24,4.24)$

The wanted probability is given by:

$P(X>16)=P(Y>15.5)$

$=P\left(\frac{Y-24}{4.24}>\frac{15.5-24}{4.24}\right)$

$=1-\varphi \left(\frac{15.5-24}{4.24}\right)$

$=1-\left(\varphi \right(-2\left)\right)$

$=1-(1-\varphi (2\left)\right)$

$=0.9772$

Part (c) Step 1: Given information

Given in the question that, Four friends, Janice, Barbara, Kathy and Roberta, decided to carpool together to get to school. Each day the driver would be chosen by randomly selecting one of the four names. They carpool to school for $96$days.

Part (c) Step 2: Explanation

Variable $X$has a binomial distribution:

$X~B(96,0.25)$

The mean:

$\mu =n·p=96·0.25=24$

Standard deviation:

$s=\sqrt{npq}=4.24$

Now random variable $Y$has a normal distribution:

$Y~N(24,4.24)$

The wanted probability is given by:

$P(X=24)=P(23.5<Y<24.5)$

$=P\left(\frac{23.5-24}{4.24}<\frac{Y-24}{4.24}<\frac{24.5-24}{4.24}\right)$

$=\varphi \left(\frac{24.5-24}{4.24}\right)-\varphi \left(\frac{23.5-24}{4.24}\right)$

$=\varphi \left(0.12\right)-\varphi (-0.12)$

$=\varphi \left(0.12\right)-(1-\varphi (0.12\left)\right)$

$=2\varphi \left(0.12\right)-1$

$=2\cdot 0.5478-1$

$=0.0956$