Question

Two 5-year girls, Alyse and Jocelyn, have been training to run a 1-mile race. Alyse's 1 mile time A is approximately Normally distributed with a mean of 13.5 and standard deviation of 2.5 minutes. Jocelyn's 1-mile time J is approximately Normally distributed with a mean of 12 minutes and a standard deviation of 1.5 minutes. Assuming A and J are independent random variables, what is the probability that Alyse has a smaller time than Jocelyn in a 1 mile race on a randomly selected day?

Answer 1

Answer:

1.7 × 10⁻⁴

Step-by-step explanation:

The question relates to a two sample z-test for the comparison between the means of the two samples  

The null hypothesis is H₀:  μ₁ ≤ μ₂

The alternative hypothesis is Hₐ: μ₁ > μ₂

$z=\dfrac{(\bar{x}_1-\bar{x}_2)-(\mu_{1}-\mu _{2} )}{\sqrt{\dfrac{\sigma_{1}^{2} }{n_{1}}-\dfrac{\sigma _{2}^{2}}{n_{2}}}}$

Where;

$\bar {x}_1$ = 13.5

$\bar {x}_2$ = 12

σ₁ = 2.5

σ₂ = 1.5

We set our α level at 0.05

Therefore, our critical z = ± 1.96

For n₁ = n₂ = 23, we have;

$z=\dfrac{(13.5-12)-(0)}{\sqrt{\dfrac{2.5^{2} }{23}-\dfrac{1.5^{2}}{23}}} = 3.5969$

We reject the null hypothesis at α = 0.05, as our z-value, 3.5969 is larger than the critical z, 1.96 or mathematically, since 3.5969 > 1.96

Therefore, there is enough statistical evidence to suggest that Alyse time is larger than Jocelyn in a 1 mile race on a randomly select day and the probability that Alyse has a larger time than Jocelyn is 0.99983

Therefore;

The probability that Alyse has a smaller time than Jocelyn is 1 - 0.99983 = 0.00017 = 1.7 × 10⁻⁴.

Answer 2

Refer the below solution for better understanding.

Step-by-step explanation:

Given :

Standard Deviation, $\rm \sigma_1 = 2.5\; and\;\sigma_2= 1.5$

Calculation :

The null Hypothesis is $\rm H_0: \mu_1\leq\mu_2$

The alternative Hypothesis is $\rm H_a: \mu_1>\mu_2$

Now,

$z = \dfrac{(\bar{x_1}-\bar{x_2})-(\mu_1-\mu_2)}{\sqrt{\dfrac{\sigma_1^2}{n_1}-\dfrac{\sigma_2^2}{n_2}} }$  --------------- (1)

putting the values of

$\bar{x_1}=13.5,\;\bar{x_2}=12,\;\sigma_1=2.5,\;\sigma_2=1.5$

in equation (1)

$z=\dfrac{(13.5-12)-(0)}{\sqrt{\dfrac{2.5^2}{23}-\dfrac{1.5^2}{23}} }$

z = 3.5969

We reject the null hypothesis at

$\alpha =0.05$

z-value is larger than the critical z.

3.5969 > 1.96

Therefore, there is enough statistical evidence to suggest that Alyse time is larger than Jocelyn in a 1 mile race on a randomly select day.

The probability that Alyse has a larger time than Jocelyn is 0.99983

Therefore, the probability that Alyse has a smaller time than Jocelyn is

$1-0.99983 = 0.00017 = 1.7\times10^-^4$

For more information, refer the link given below

https://brainly.com/question/20165896?referrer=searchResults